Home of Science

Physics Calculations

Tag: physics calculations

Resistivity Formula, Definition, Unit and Calculations

What is Resistivity?

Resistivity Formula and Definition: Resistivity is a physical property of materials that describes their ability to resist the flow of electric current. The si unit of resistivity is in ohm-meters (Ωm). It is determined by the material’s length, Area, and resistance or temperature.

Video Explanation of Resisitivity and Conductivity with Solved Problems

Materials with high resistivity impede the flow of electric current more than materials with low resistivity. Resistivity is an important property in the design and use of electrical and electronic components and systems.

resistivity formula

The above picture describes the formula we use to calculate resistivity. We will now discuss the resistivity formula in the section below.

Read: Resistance vs Resistivity

Resistivity Formula

Resistivity Formula
Picture: Resistivity Formula

We can use any of the three resistivity formula below to solve a problem depending on the question available to us:

  1. ρ = (R x A) / L,

or

2. ρ = (Rπr2) / L,

or

3. ρ = (Rπd2) / 4L,

Where

ρ = Resistivity

R = Resistance

A = Area

L = Length

r = radius of a wire in meters square

d = diameter of a wire in meters square

Electrical Conductivity Formula

Definition of Electrical Conductivity: Electrical conductivity is the reciprocal of resistivity. The si unit of per ohm-meters (Ωm)-1 or Ω-1m-1.

The electrical conductivity formula is: σ = 1 / ρ = (L / RA) = L / (Rπr2) = 4L / (Rπd2)

Therefore, you can use any of the above formulas to calculate conductivity.

Additionally, you can simply find the resistivity and then divide it by one (1) to obtain the conductivity of the material.

Points to Note While Calculating Resistivity or Conductivity

Always make sure you do not your calculations in uniform units of measurement. Avoid mixing up m and cm or m and mm. You need to apply your knowledge of prefixes in physics.

Therefore, its preferable to express all measurements in si units.

You may also like to read:

What are Prefixes in Physics

Resistivity Formula: Practice Problems

Here are a few solved problems on resistivity:

Problem 1

Calculate the resistivity of a wire length 4m and cross-sectional area 4×10-7m2 if its resistance is 3.0 ohms.

[UTME 2019]

Video of Solved Problem

Solution

Data: To calculate the resistivity of a wire, the first step is to bring out the data from the question above

As we can see,

Length of the wire (L) = 4m

Cross-sectional Area (A) = 4×10-7m2

Resistance of the wire (R) = 3Ω

Resistivity of the wire (ρ) = ?

The formula is given as

Resistivity formula: ρ = (RA) / L

Therefore, by substituting the values of ρ, R and L into the above equation, we will have

ρ = ( 3 x 4×10-7) / 4

The above expression will give us

ρ = 0.0000012 / 4

Therefore, the resistivity of a wire is

ρ = 0.0000003

ρ = 3×10-7Ωm

Hence, the resistivity of the wire is 3×10-7Ωm

Problem 2

What is the resistivity of a cylindrical material wire whose 1.0 meters length has a resistance of 2.0 ohms? The diameter of the wire is 0.5 millimetres.

Answer

The final answer to the above question is 3.93 x 10-7 ohm – meter

Resistivity
Resistivity Calculation

Explanation

L = 1.0 m

r = d/2 = 0.5 / 2 = 0.25 mm

We can now covert r into meters

r = 0.25 mm = 0.25/1000 = 2.5 x 10-4 m

A = πr2 = 3.142 x ( 2.5 x 10-4)2

Since the resistivity formula is ρ = (R x A) / L

ρ = (2 x 3.142 x 2.5 x 10-4)2 / 1

ρ = 3.93 x 10-7 Ωm

Therefore, the resistivity of the cylindrical material is 3.93 x 10-7 ohm – meter

Problem 3

A wire of length 100 cm and cross-sectional area 2.0 x 10-3 cm2 has a resistance of 0.10 Ω. Calculate its electrical conductivity.

Conductivity Calculation
Conductivity Calculation

Solution

Length (L) = 100 cm, cross-sectional area (A)= 2.0 x 10-3 cm2, and Resistance (R) = 0.10 Ω

Applying the formula for conductivity, we will have:

σ = 1 / ρ = (100 cm / 0.10 Ω x 2.0 x 10-3cm2) = 5 x 105 Ω-1cm-1

Therefore, the electrical conductivity of the wire is 5 x 105 Ω-1cm-1

Problem 4

A constantan wire has a cross-sectional area of 4 x 10-8 m2 and a resisitivity of 1.1 x 10-6 Ωm. If a resisitor of resistance 11 Ω is to be made from this wire, calculate the length of the wire required.

Solution

A = 4 x 10-8 m2, ρ = 1.1 x 10-6 Ωm, and R = 11 Ω

We are to find L

Formula: L = RA / ρ = (11 x 4 x 10-8) / 1.1 x 10-6 = 0.4 m

Problem 5

A 2 meters wire of resistivity 5.5 x 10-7Ωm has a cross-sectional area of 0.50 mm2. Calculate its resistance.

Solution

L = 2 m, ρ = 5.5 x 10-7Ωm, and A = 0.50 mm2 = 0.5 x 10-6 m (1 mm2 = 1 x 10-6 m)

We are to find R

Formula: R = ρL / A = (5.5 x 10-7 x 2) / 0.5 x 10-6 = 2.2 Ω

Therefore, the resistance of the wire is 2.2 ohms.

Problem 6

A wire of length 15 m made of a material of resistivity 1.8 x 10-6 Ωm has a resistance of 0.27Ω. Determine the area of the wire.

Solution

A = ρL / R = (1.8 x 10-6 x 15) / 0.27 = 1.0 x 10-4 m2.

Problem 7

A wire of 5Ω resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remainsthe same and the cross-sectional area is halved, what will be the new resistance?

Solution

Original Wire – R1 = 5Ω, L1 = L, ρ1 = ρ, and A1 = A

Formula for the original wire => ρ1 = R1A1 / L1

New Wire – R2 = ?, L2 = 2L (twice the original length), ρ2 = ρ (resistivity remains the same), A = A/2 (cross-secrtional area is halved)

Formula for the new wire => ρ2 = R2A2 / L2

When we equate the original wire with the new wire, we will now obtain

ρ1 = ρ2

Therefore,

R1A1 / L1 = R2A2 / L2

Thus, we will now make R2 subject of the formula

R2 = R1A1L2 / A2L1

Putting the values of our data into the above equation, we will now have:

R2 = (5 x A x 2L) / ((A/2) x L) = (5 x 2 x A x 2L) / (A x L) = 5 x 2 x 2 = 20Ω

Therefore, the new resistance (R2) is 20 Ohms.

Resistivity vs Conducitivity

Below is a table summarizing the main differences between resistivity and conductivity:

PropertyResistivityConductivity
DefinitionA material’s resistance to the flow of electric current.A material’s ability to conduct electric current.
Symbolρ (rho)σ (sigma)
UnitOhm-meters (Ω·m)Siemens per meter (S/m) or mhos per meter (mho/m)
MeasurementIndirectly measured using a resistivity meter or ohmmeter.Directly measured using a conductivity meter.
RelationshipInversely related to conductivity: σ = 1/ρ or ρ = 1/σ.Inversely related to resistivity: σ = 1/ρ or ρ = 1/σ.
Material TypesInsulators have high resistivity.Metals have high conductivity.
Temperature EffectIn most materials, resistivity increases with temperature.In most materials, conductivity decreases with temperature.
Role in Electrical EngineeringCrucial for analyzing and designing electrical circuits.Important for selecting suitable materials for electrical components.
Influence on Wire ThicknessLower resistivity materials require thicker wires for the same current.Higher conductivity materials can use thinner wires for the same current.
Impurity EffectImpurities can increase resistivity.Impurities can decrease conductivity.
Resistivity vs Conductivity

Please note that resistivity and conductivity are mathematically related, and as one property increases, the other decreases, and vice versa. This table above highlights their key differences. Therefore, it makes it easier to understand their distinct roles and characteristics.

Resistivity vs Resistance

Here is another table summarizing the main differences between resistivity and resistance:

PropertyResistivityResistance
DefinitionIntrinsic property of a material that describes how strongly it resists the flow of electric current.Property of an object or material that measures the opposition to the flow of electric current.
Symbolρ (rho)R
UnitOhm-meters (Ω·m)Ohms (Ω)
MeasurementMeasured for materials.Measured for objects or components.
Dependent onMaterial composition, temperature, and impurities.Material’s resistivity and its physical dimensions (length and cross-sectional area).
RelationshipRelated to conductivity: σ = 1/ρ or ρ = 1/σ.Related to resistivity: R = ρ * (L / A), where “L” is length and “A” is cross-sectional area.
ApplicationUsed to analyze material behavior under electric fields.Used to analyze and design electrical circuits and systems.
ExamplesDifferent materials have different resistivity values.Components like resistors have specific resistance values.
Temperature EffectResistivity may increase or decrease with temperature, depending on the material type.Resistance typically increases with temperature in conductors. In some cases, it may decrease in certain semiconductors.
Material PropertyAn intrinsic property of the material itself.Depends on both the material and its physical dimensions.
FormulaNo direct formula; related to conductivity.R = ρ * (L / A)
Resistivity vs Resistance

The table above highlights the key distinctions between resistivity, which is a material property, and resistance, which depends on both the material and its dimensions. Therefore, as resistivity describes how a material resists electric current flow, resistance quantifies the opposition to current flow in a specific object or component.

Frequently Asked Questions about Resistivity and Conductivity

Q1: What is Resistivity?

Resistivity, denoted by the symbol “ρ” (rho), is a fundamental property of a material that describes how strongly it resists the flow of electric current. It is an intrinsic characteristic of a material, independent of its shape or size. Resistivity is measured in ohm-meters (Ω·m) and is a crucial parameter in determining the electrical behavior of various materials.

Q2: What is Conductivity?

Conductivity is the reciprocal of resistivity and is represented by the symbol “σ” (sigma). It measures the ability of a material to conduct electric current. The higher the conductivity of a material, the better it can conduct electricity. Conductivity is also measured in siemens per meter (S/m) or mhos per meter (mho/m).

Q3: How are Resistivity and Conductivity related to each other?

Resistivity and conductivity are mathematically related through the equation: σ = 1/ρ or ρ = 1/σ. As one property increases, the other decreases, and vice versa. Highly conductive materials have low resistivity, while materials with high resistivity have low conductivity.

Q4: How does Temperature affect Resistivity and Conductivity?

In most materials, as the temperature increases, the resistivity also increases. This means that the conductivity decreases with rising temperature. However, in some materials known as “positive temperature coefficient” materials, the resistivity decreases with increasing temperature, leading to an increase in conductivity.

Q5: What are the units of Resistivity and Conductivity?

The units of resistivity are ohm-meters (Ω·m), while the units of conductivity are siemens per meter (S/m) or mhos per meter (mho/m).

Q6: What are some examples of materials with high Resistivity?

Materials like rubber, glass, ceramics, and most insulators have high resistivity. These materials are generally poor conductors of electricity due to the presence of tightly bound electrons, which hinder the flow of current

Q7: What are some examples of materials with high Conductivity?

Metals, such as copper, aluminum, silver, and gold, exhibit high conductivity due to their abundance of free electrons that can easily carry electric charges.

Q8: How is Resistivity different from Resistance?

Resistivity is an intrinsic property of a material, while resistance depends on both the resistivity of the material and its physical dimensions. Resistance (R) is given by the equation R = ρ * (L / A), where “L” is the length and “A” is the cross-sectional area of the material.

Q9: What factors influence the Resistivity of a material?

The resistivity of a material depends on its composition, temperature, and the presence of impurities. Crystalline structures tend to have lower resistivity than amorphous materials.

Q10: Why is the concept of Resistivity important in Electrical Engineering?

Resistivity plays a crucial role in the design and analysis of electrical circuits and systems. It helps engineers understand how different materials will behave under the flow of electric current, allowing them to choose the right materials for specific applications.

Q11: How is Resistivity measured experimentally?

Resistivity is measured using a device called a “resistivity meter” or “ohmmeter.” The measurement involves passing a known current through a sample of the material and measuring the voltage drop across it. The resistivity is then calculated using the dimensions of the sample.

Q12: Can Resistivity be used to predict a material’s behavior in an electric field?

Yes, resistivity is an essential parameter for predicting how a material will respond to an electric field. Materials with higher resistivity will experience more significant voltage drops and dissipate more power, whereas highly conductive materials will carry current with minimal losses.

Q13: How does Moisture affect the Resistivity of materials?

Moisture can significantly alter the resistivity of certain materials, particularly insulators. Water, being a good conductor, can create paths for current flow, reducing the overall resistivity of the material. This effect can lead to electrical failures in insulating materials in humid environments.

Q14: Are there materials with variable Resistivity?

Yes, certain materials exhibit a property called “variable resistivity” or “resistive switching.” These materials can change their resistivity under the influence of an electric field, allowing them to be used in devices like resistive random-access memory (RRAM) and memristors.

Q15: What is the role of Resistivity in determining the thickness of electrical wires?

Resistivity is taken into consideration when determining the appropriate thickness of electrical wires. Lower resistivity materials require thinner wires to achieve the desired current-carrying capacity without significant power losses, while high resistivity materials necessitate thicker wires for the same purpose.

I hope this article has helped you understand the general topic on resistivity definition, resistivity formula, resistivity units, and resistivity calculations. To simplify the topic, I added conductivity definition, conductivity unit, conductivity formula, and conductivity calculations to broaden your understanding. I wish you goodluck!

You may also like to read:

General Gas Law in Physics

Reference

Relation Between Resistance and Length

What is Velocity Units in Physics

What are velocity Units in Physics?

The accepted velocity unit in physics is in meters per second. and the symbol is m/s. Other units are miles per hour (mph), kilometres per hour (km/h), knot, feet per second, and inch per second. We use these other units according to the industry and geographic locations. You may wish to continue reading to know more about all the units of velocity in physics.

Velocity Units in Physics

Are you familiar with Velocity units in Physics? Velocity, in simple terms, refers to the speed of an object in a specified direction or more like a translational motion. It is an important concept in physics and engineering that helps us understand the movement of objects in the world around us.

Velocity is a fundamental concept in physics and engineering. We use it in a wide range of applications, from calculating the speed of a car to understanding the motion of planets in space. To measure velocity, we use velocity units, which are the measurement units that help us to express the speed of an object.

Velocity units are the measurement units that we use to describe the speed of an object. In this article, we will discuss the different types of velocity units, their conversion factors, and how we use them in calculations. We will also answer some frequently asked questions about Velocity units to help you better understand this critical concept.

Understanding the Basics of Velocity Units

Velocity units simply describe the speed of an object in a given direction. The most common units are meters per second (m/s), miles per hour (mph), and kilometres per hour (km/h). These units are used to show the speed of objects in different contexts.

For example, we know that the velocity formula is displacement over time (v=x/t). Additionally, the unit of displacement(x) in physics is in meters(m), while the unit of time(t) is in seconds(s). Therefore, by looking at the symbols that represent, velocity as v, displacement as x, and time as t. Where the symbols of meters is m, and the second is s. We will see that

Velocity(v) = displacement(x) / time(t)

v = x/t

If we are to use the symbols for the units of x and t, we will have:

v = m/s

Thus, its clear that

velocity = meters / second

Therefore, we can clearly see that the s.i unit for velocity (v) is in meters per second (m/s) which is the unit of measurement for velocity.

1. Meters per Second (m/s)

Meters per second (m/s) is the most commonly used velocity unit in the field of physics and engineering. We define it as the distance (in a specific direction) travelled in meters in one second. This unit describes the speed of objects that move at high speeds, such as rockets and aeroplanes.

2. Miles per Hour (mph)

Miles per hour (mph) is a very familiar unit in the United States and the United Kingdom. It is the distance (in a specific direction) travelled in miles in one hour. This unit helps us to understand the speed of cars, aeroplanes, and other objects that move at moderate speeds.

3. Kilometres per Hour (km/h)

Kilometres per hour (km/h) is very common in the metric system. We see it as the distance travelled in kilometres in one hour. This unit expresses the speed of cars, aeroplanes, and other objects that move at moderate speeds.

Velocity Units Conversion Factors

Converting between the units of velocity is important in physics and engineering. This is because different applications may require different units. Here are some of the conversion factors for velocity units:

  • 1 meter per second (m/s) = 2.23694 miles per hour (mph)
  • 1 mile per hour (mph) = 1.60934 kilometres per hour (km/h)
  • 1 kilometre per hour (km/h) = 0.621371 miles per hour (mph)
  • 1 kilometre per hour (km/h) = 0.277778 meter per second (m/s)
  • 1 meter per second (m/s) = 3.28084 foot per second (ft/s)

How We Use Velocity Units in Calculations

We use velocity units in a wide range of calculations in physics and engineering. Here are some of the velocity units are that we use in calculations:

1. Speed Calculation

We use velocity units to calculate the speed of an object. The formula for calculating speed is:

  • Speed = Distance / Time = Meters / Second

Velocity unit shows the distance and time in the equation. For example, we measure the distance in meters and the time is in seconds. Thus, the velocity unit would be meters per second (m/s).

2. Acceleration Calculation

Velocity units also helps us to calculate acceleration. Acceleration is the rate at which an object changes its velocity. The formula for calculating acceleration is:

  • Acceleration = (Final Velocity – Initial Velocity) / Time

Velocity units in acceleration are the velocity and time in the equation.

3. Force Calculation

We also use velocity units in calculating force. Force is the product of mass and acceleration. Additionally, we measure it in Newtons (N). The formula for calculating force is:

  • Force = Mass x Acceleration

For example, if we can measure the acceleration in meters per second squared (ms-2). The Velocity unit would be meters per second (m/s).

Energy Calculation

Velocity units are also important in energy calculations. Energy is the product of force and distance. Th unit of measurement of energy is in Joules (J). The formula for calculating energy is:

  • Energy = Force x Distance

Velocity units also expresses the force and distance in the equation. For example, if the measurement of force is in Newtons (N), and the distance is in meters (m). The Velocity unit would be meters per second (m/s).

Frequently Asked Questions

  1. What is the difference between speed and Velocity?
  • Speed is a scalar quantity that refers to the rate at which an object moves, while Velocity is a vector quantity that refers to the rate at which an object moves in a particular direction.
  1. What is the SI unit of Velocity?
  • The SI unit of Velocity is meters per second (m/s).
  1. How do you convert Velocity units?
  • You can convert Velocity units using conversion factors. For example, to convert meters per second (m/s) to miles per hour (mph), you can multiply the value in m/s by 2.23694.
  1. What is the formula for calculating speed?
  • The formula for calculating speed is: Speed = Distance / Time
  1. What is the formula for calculating acceleration?
  • The formula for calculating acceleration is: Acceleration = (Final Velocity – Initial Velocity) / Time
  1. What is the formula for calculating force?
  • The formula for calculating force is: Force = Mass x Acceleration

Conclusion

Velocity units are very important in physics and engineering. They help us understand objects’ movement in the world around us. By understanding Velocity units, you can calculate the speed, acceleration, force, and energy of objects in different contexts.

Remember, Velocity units are just one part of the larger world of physics and engineering. By continuing to learn and explore, you can gain a deeper understanding of the principles that govern the movement of objects in the world around us.

You may also like to read:

What is Velocity Formula in Physics?

Newton’s Laws of Motion

Newton’s Three Laws of Motion

Over the last 300 years, Newton’s laws of motion became one of the most famous laws of physics. The laws were stated by a famous English physicist and mathematician, and his name is Sir Isaac Newton. Below is a video explaining of the three laws of motion.

Newton’s three laws of motion are as follows:

  1. A body will continue to be at rest or in a motion until it’s acted upon by an external force. This is also known as the law of inertia (“Inertia of a body is its reluctance to start moving when at rest or stationary and its reluctance to stop moving once it is moving”).  
  2. The rate of change of momentum is proportional to the external force and takes place in the direction of the applied force.
  3. When an external force is applied to a body, there is always an equal and opposite reaction. The law can also be defined as to every action, there is an equal and opposite reaction.
Newtons Laws of Motion
Newton’s Laws of Motion

You may also like to read:

Newton’s Laws of Motion Project Ideas

Brief History of the Laws of Motion

Newton was born in England on 4th of January, 1643 in Woolsthorpe Manor House. During his time, children were trained to rear sheeps to earn a living. However, Newton was uninterested in making a living as a farmer. Instead, he was more attracted to books. 

At young age, he found love in the books that were arranged in the shelves of his parent’s library. The famous scientist dedicated his time to observing the possible occurrences that happen between sunrise to sunset, spring to summer, autumn to winter. Newton tried to understand what triggers these changes, and it led him to understanding the gravity. 

He observed the effect of sun, wind and water on materials like water clocks, windmills, and kites. Isaac finally realizes that the earth and the sky work on the same principle which is contrary to Aristotle’s understanding.

The influential English scientist devoted himself to reading the works of ancient Greeks, Islamic and medieval views on gravity. Newton asked himself as to why an apple fruit falls on the ground instead of going up to the sky. He provided mathematical evidence to Galilei Galileo and Johannes Kepler’s theory of gravity.   

Isaac Newton started his journey of research at the age of 23. He came up with a book on classical mechanics, and the name of the book is “Principia Mathematica Philosophiae Naturalis.”.  The meaning of the book in English is “Mathematical principles of natural philosophy”. 

This book perfectly explained the 3 laws of motion. The three (3) laws of motion shaped our understanding of gravity and we call them Newton’s laws of motion.

Newton’s Laws of Motion in our Everyday  Life

We apply Newton’s laws of motion in our everyday life. For example, while playing soccer, we must apply force for the ball to move. Once we apply force on the ball, it will head towards our target due to momentum. The wide range of applications of Newton’s laws is as follows:

  • Basketball
  • Rocket science
  • Javelin throw
  • Table tennis
  • Squash sport 
  • Riding Bicycle
  • Driving a Bus on the road
  • Eating with a Spoon
  • An airplane taking off
  • A bag resting on a desk, and many more. 

You may also like to read:

 The Newton’s Second Law Practice Problems

Second Law of Motion Examples

Newton’s First (1st) Law of Motion: Law of Inertia

The first law of motion states that A body will continue to be at rest or in motion, until it’s acted upon by an external force. According to this law, when you hide an object in your room. The object will continue to remain in that place for millions of years, until someone or something applies a force to move the object from its original hiding position.

Therefore, when you keep something in a safe place. We can easily say that the object will remain in that place until you decide to change its position or someone moves it from its original position. 

Additionally, the same law state that when you set your car in motion. The car will continue to move in a straight line and will never stop. It will wait for you to apply brakes before it can stop. Additionally, the car may also stop if the engine is bad or the fuel is empty. Hence, there must be an application of external force for the car to stop. 

Also, if you don’t control the steering, it will continue to move in a straight line until you decide to go left or right. This scenario also explains Newton’s first law of motion or law of inertia.

Understanding Inertia

Newton’s first law of motion is also known as the law of inertia. As defined in his book, the law of inertia is the power by which an object, if it’s at rest, stays at rest, or if in motion, will travel in a straight line.

To understand this law, try rolling a marble on smooth tiles, the marble will continue to move until an object blocks its movement. Alternatively, the friction of the tiles will make the speed of the marble keep decreasing until it comes to rest.

This scenario of a rolling marble against the tiles perfectly describes Newton’s first law of motion. This law explains why your coffee remains peacefully in its cup until you decide to stir it or why your car continues down the highway unless you hit the brakes.

Everyday Examples

Consider a few everyday examples:

  • A hockey puck sliding on an ice rink with no friction would continue forever if not for external forces like air resistance or the rink’s boards.
  • In the vacuum of space, where there is no air resistance, celestial bodies such as planets and moons continue in their orbits with the same speed and direction due to their inertia.

You may also like to read:

 Second Law Practice Problems

Laws of Motion Project Ideas

Newton’s Second (2nd) Law of Motion  

The second law of motion states that when you apply an external force on an object the rate of change of momentum is proportional to the applied force and it takes place in the direction of the force.

Here is the best way to explain this law, when you push a shopping cart in the supermarket. The shopping cart will move because of the external force you apply on it, and the mass of the cart.

Therefore, the mass of the , mass of the body and the external force acting upon it plays a crucial role in setting a body into motion. 

Here is another example, before buying anything, push the cart for a while. You will notice that it moves faster. Now, add some of the items you bought from the store into the cart. What did you notice? It’s heavier than it was, right? You may need to add extra force to push it this time around. 

Ask yourself again, what brought this sudden change? Here is the answer to your question: “The more mass you add to the cart, the heavier it becomes, and the more force you will need to apply to move the cart”.  

Hence, there exists a relationship between the mass of the body and the applied external force.

The mathematical expression (formula) for newton’s second law of motion is

Force (F) = Mass (m) x Acceleration (a) 

Which can be written as 

F = ma

The S.I unit of force is called Newton (Honors Isaac Newton) or kilogram meter per second square. The symbol of Newton is N, while the symbol for kilogram meters  per second square is kgm/s2

The unit of mass is in kilograms (kg)

The unit of acceleration is in meter per second square (m/s2)

Practical Examples

In simpler terms, the more force you apply to an object, the faster it will accelerate, while a heavier object requires more force to achieve the same acceleration.

  • When you push a shopping cart, the harder you push, the faster it accelerates.
  • In sports, athletes understand this law intuitively. A soccer player applies more force to the ball to make it accelerate quickly.

How to Derive Newton’s Second Law

According to Newton’s second law, the rate of change of momentum is proportional to the applied force. Therefore, this law can be translated into 

Force (F) = change in momentum / time (t)

Change in momentum is the product of mass and velocity.

Thus,

Change in momentum = mass x velocity

Which can be further broken down into

Change in momentum = mv – mu   [where v = final velocity and u = initial velocity]

Therefore, we can insert (Change in momentum = mv – mu) into [Force = change in momentum / time] to obtain

F =  (mv – mu)  / t

Hence, 

F = m (v – u) / t 

And acceleration is the rate of change of displacement

Acceleration, a = (v – u) / t

And we have F = ma which is Newton’s second law of motion. 

Momentum 

This is the product mass and velocity of an object. The S.I unit of momentum is kilogram meter per second (kgm/s). The formula for calculating momentum is 

Momentum (M) = mass (m) x velocity (v)

The rate of change of momentum = (change in momentum) / time

Impulse of a Force

The impulse of a force is the instantaneous response of an object to the force that acts on it. We can also define impulse (I) as the product of the force and the time it  takes momentum to change. The S.I unit of impulse is Newton-second (Ns).

Impulse (I) = Force (F) x Time (t)

Calculations Involving Second Law of Motion

Here are some questions to help you understand the applications of Newton’s second law of motion to solve a problem.

Problem 1

An object of mass 7.5 kilograms has a force applied to it that pulls eastward with a magnitude of 50 Newtons and another applied to it that pulls westward with a magnitude of 150 Newtons. What is the westward acceleration of the object? Give your answer to one decimal place.

Data

Mass of the object is m = 7.5 N

Initial Force, F1 = 50 Newtons

Final force, F2 = 150 Newtons

Unknown

Acceleration of the object, a = ?

Formula

The formula we need to apply to solve this problem is

FNet = ma

But FNet = F2 – F1

Therefore, w can further break it down into

F2 – F1 = ma   

Solution

We can now insert our data into F2 – F1 = ma   

150 – 50 = 7.5 x a

Make a subject of the formula

a = (150 – 50) / 7.5

Hence,

a = 100/7.5 = 13 m/s2

Problem 2

An object of mass 3 kilograms has a force of 9 Newtons applied to it. At what rate does the force accelerate the object? 

Data

Mass of the object, m = 3 kg

Applied force, F = 9 N

Unknown

Acceleration, a = ?

Formula

The formula that can help us to solve this problem is

a = F/m

Solution

We can substitute our formula with our data

Acceleration, a = 9/3 = 3 m/s2

Therefore, the acceleration (a) is 3 meters per second square.

Problem 3

A boy pushes his toy car with a cat in inside. The mass of the toy car and the cat put together is 70 kilograms. The toy car accelerates at 0.9 meters per second square. Find the force that the boy applies to pull the toy car and the cat together.

Data

Mass of the toy car with a cat is m = 70 kg

Acceleration of the toy car, a = 0.9 m/s2

Unknown

Force, F = ?

Formula

We will use the formula F = ma to solve the problem

Solution

Insert the data into the above formula

F = 70 x 0.9 = 63 N

Therefore, the applied force is 63 Newtons

Third Newton’s Law of Motion

The third Newton’s law of motion states that action and reaction are equal and opposite. This law explains that when a force is applied to a body, there is a chain of equal and opposite reactions that will follow.

For example, when you blow a balloon, and later release the air entrance. The balloon will run away from you. 

What if we decide to insert the balloon inside a toy canoe? We will blow the balloon and position its air entrance to face the opposite direction of the canoe’s motion. After releasing the air entrance of the balloon, it will set the toy canoe in motion on water.

Therefore, we can see that when an object X applies a force on another object Y. Object Y will equally apply the same type of force on X. Both applied forces are equal in magnitude but opposite in direction. 

We can demonstrate this mathematically as 

Force X = – Force Y  [ Force X is the action force, and Force Y is the reaction Force]

Another example of Newton’s third law of motion is how the rocket operates. When a rocket is launched into space, it will accelerate away from the earth surface. This is thanks to the burning fuel of the rocket that kept igniting fire at the back of the rocket to push it in the upward direction.

We have many examples to demonstrate the effect of this law in our everyday life. Among these examples are: 

  • The lift of an airplane
  • Student’s school bag on the desk
  • A hammer hitting a nail
  • A bird flying
  • A boy pushing against the wall
  • Applying brakes on your car
  • A duck swimming in the water 

Everyday Examples

In other words, when one object exerts a force on another object, the second object exerts an equal and opposite force on the first.

  • Walking: When you walk, your foot pushes backward on the ground, and the ground pushes your foot forward, propelling you forward.
  • Rocket Propulsion: Rockets work by expelling high-speed exhaust gases backward. The force of the exhaust gas going backward propels the rocket forward.

Principle of Conservation of Momentum

We can state the principle of conservation of momentum in three ways:

  1. The principle states that when two bodies that are traveling in a straight line collide, the total force before collision is equal to the total force after collision provided no external force acts on the body. 
  2. In any system of colliding bodies, the total momentum is always conserved provided that there is no net external force acting on the system.  
  3. The total momentum of an isolated or closed system of colliding bodies remains constant.

Therefore, the mathematical definition of this principle is

m1v1 = m2v2 [Where m1v1 is the initial momentum, and m2v2 is the final momentum

Now, if we have two balls M and N exerting equal forces on each other. We can say that

FM = – FN

And since F = ma

Then, it’s easy to say that 

mMaM = – mNaN

Additionally, a = (v – u) / t

Which implies

mM [ (vm – um) / t ] = – mN [ (vN – uN) / t ]

And we will end up with

mMvm – mMum = – ( mNvN – mNuN)

The above equation will give us

mMum + mNuN = mMvm + mNuN

And the above expression shows that momentum before collision is equal to the momentum after collision.

Applications of Newton’s Third Law to Solve a Problem

Here are a few problems to help you understand how to solve problems that are related to newtons third law of motion, momentum, and impulse.

Problem 1

A body (P) has mass of 5 kilograms moving with a velocity of 30 m/s collides with another body (Q), moving in opposite direction with a velocity of 20 m/s. If both bodies now move in the direction of P at a velocity of 10 m/s. Calculate the mass of Q.

Newton’s Laws of Motion
Solution

To solve this problem, we need to remember that

m1v1 = m2v2

Which implies that

(m1 + m2)v = m1u1 – m2u2

and m1u1 – m2u2 = Before collision

(m1 + m2)v = After collision

Therefore, since m1 = 5kg, u1 = 30 m/s, m2 = ? , u2 = 20 m/s, v = 10 m/s

We can now insert the above information into (m1 + m2)v = m1u1 – m2u2 to obtain

(5 + m2) x 10 = (5 x 30) – (m2 x 20)

We will now have

50 + 10m2 = 150 – 20m2

We can now collect like terms to get

150 – 50 = 20m2 + 10m2

And it will become

100 = 30m2

By making m2 subject of the formula, we will have

m2 = 100 / 30 = 3 kg

Therefore, the mass of Q is 3 kilograms.

Problem 2

A bullet of mass 0.05 kg is fired horizontally into a 10 kg block which is fired free to move. If both bullet block move with velocity 0.5 m/s after the impact, find the velocity with which the bullet hit the body.

Data

Mass of the bullet = 0.05 kg, mass of the block = 10 kg, and the mass of both bullet and the block = 10 + 0.05 = 10.05

Initial mass of the bullet = u

The momentum before collision = Momentum after collision

0.05 x u = 10.05 x 0.5

Hence,

u = 5.025 / 0.05 = 100.5 = 101 m/s

Problem 3

A rifle of mass 15 kilograms fires a bullet of mass 60 grams with a velocity of 200 m/s. Calculate the recoil velocity of the rifle.

Solution

m1 = 15 kg, and m2 = 60 g = (60/1000) kg = 0.06 kg,

v1 = ? and v2 = 200 m/s

We need to remember that Momentum before impact = Momentum after impact

Momentum before impact = 0

And the Momentum after impact = m1 v1 + m2 v2

Hence,

m1 v1 + m2 v2 = 0

By inserting our data, we will have

15 x v1 + 0.06 x 200 = 0

Which will become

15 v1 + 12 = 0

After making v1 subject of the formula

v1 = – (12/8) = – 0.8 m/s

Therefore, the recoil velocity is 0.8 meters per second

Isaac Newton Defined the Following Quantities in His Book

Matter: Anything that takes up space

Mass: The measure of a quantity or amount of matter

Momentum: The quantity of motion, which is the product of velocity (speed) and mass

Inertia: The power by which an object, if it is at rest, stays at rest, or if in motion, will travel in a straight line

Force: An action applied upon a body

Centripetal Force: An attraction toward the center of something (as in gravity)

We apply Newton’s laws of motion in our everyday activities.

Importance and Applications

Now that we’ve explored the laws themselves, let’s consider their practical applications in our everyday lives and various fields of science and technology.

Transportation and Engineering

Newton’s Laws are fundamental in the design of vehicles and structures. They influence:

  • Automobiles: Every time you hit the gas pedal or apply the brakes, you’re experiencing Newton’s Laws in action. The force you apply determines the acceleration or deceleration of your vehicle.
  • Aviation: In aviation, understanding the interplay of forces is essential for safe and efficient flight. Pilots must balance thrust, lift, drag, and gravity to keep an aircraft in equilibrium.
  • Maritime Travel: Ships, too, rely on Newton’s Laws. When a ship’s engine propels it forward, the reaction is a force that moves the ship in the opposite direction.

Sports and Athletics

Newton’s Laws govern the dynamics of sports and athletics:

  • Baseball: When a pitcher throws a fastball, the force applied to the ball determines its speed and trajectory. Hitters use this knowledge to time their swings.
  • Swimming: Swimmers maximize their efficiency by minimizing water resistance and using the principles of buoyancy and propulsion, all in line with Newton’s Laws.
  • Billiards: In billiards, players use the principles of angles and collisions to control the path of the balls.

Space Exploration and Astronomy

Newton’s Laws are integral to space exploration. They underpin:

  • Spacecraft: Rockets and spacecraft rely on these laws to propel them through the vacuum of space. Rocket engines function by expelling mass at high velocities, creating a reaction that propels the spacecraft in the opposite direction. Without Newton’s Third Law, space exploration as we know it would be impossible.

Historical Significance

The importance of Newton’s Laws is not limited to their practical applications. They also hold historical significance as they mark a turning point in the world of science.

A Scientific Revolution

Newton’s Laws were part of a broader scientific revolution that also included the works of Galileo, Copernicus, and Kepler. They offered a unified and mathematically rigorous framework for understanding motion and paved the way for classical physics.

Revolutionizing Astronomy

In addition to their impact on physics, Newton’s Laws played a pivotal role in the field of astronomy. His law of universal gravitation, which describes how all objects with mass are attracted to each other, led to the understanding of planetary motion and orbits. It was Newton’s mathematical genius that allowed scientists to predict the movement of celestial bodies with unparalleled accuracy.

Educational Impact

Newton’s Laws play a crucial role in science education and the development of critical thinking skills.

Importance of Teaching

These laws are fundamental to STEM (Science, Technology, Engineering, and Mathematics) education. They introduce students to the principles of physics and help them develop critical thinking skills. Educators emphasize the importance of teaching Newton’s Laws as a gateway to understanding the scientific method.

Resources for Learning

Numerous educational resources and tools are available to facilitate the learning of Newton’s Laws. These include interactive simulations, video lessons, and hands-on experiments that make physics accessible and engaging for students of all ages.

Newton’s Laws and the Bigger Picture

Newton’s Laws are not isolated principles; they are interconnected with other scientific concepts.

Gravity and Universal Gravitation

Newton’s Laws and the law of universal gravitation are closely related. Gravity, the force of attraction between masses, adheres to the principles set forth in Newton’s Laws. The orbits of celestial bodies, planetary motion, and the behavior of objects in free fall are all governed by a combination of these laws.

Laws of Thermodynamics

While the laws of thermodynamics primarily describe the flow of heat and energy, they are not separate from Newton’s Laws. For example, the behavior of gas particles in a container can be explained using the laws of motion as molecules collide and exert forces on one another.

Quantum Mechanics

Even in the realm of quantum mechanics, which explores the behavior of particles on the smallest scales, the foundations laid by Newton’s Laws still apply. While quantum mechanics introduces new principles, the motion of particles and the forces acting upon them remain consistent with Newton’s insights.

Quotes and Insights from Experts

To highlight the enduring importance of Newton’s Laws, let’s turn to the insights of renowned physicists and scientists.

Albert Einstein

Albert Einstein, the legendary physicist, once remarked, “You have this elementary particle, the electron, and it is attracted by positive charge, and repelled by negative charge. Newton would have despised that theory.” Einstein’s comment underscores the profound impact of Newton’s work and the subsequent developments in physics.

Richard Feynman

Richard Feynman, another Nobel laureate in physics, praised Newton’s contributions by stating, “In his day, the great truths which now seem so obvious were unknown… He took the known facts and extended our ideas to a remarkable degree in a precise quantitative way.” Feynman recognized the precision and depth of Newton’s work, which laid the groundwork for future scientific discoveries.

Interactive Demonstrations and Simulations

To truly grasp the significance of Newton’s Laws, there’s no substitute for hands-on experience. Interactive simulations and demonstrations can bring these laws to life, enabling learners to experiment and observe the principles of motion.

Hands-On Learning

We recommend exploring online platforms, such as PhET Interactive Simulations, where you can interact with virtual experiments that demonstrate the concepts of force, motion, and equilibrium. These tools offer an intuitive and engaging way to gain a deeper understanding of Newton’s Laws.

Conclusion

In conclusion, the importance of Newton’s Laws of Motion cannot be overstated. These fundamental principles, introduced by the genius of Sir Isaac Newton, continue to shape our understanding of the physical world. They govern everything from the way we travel to the devices we use, from the structures we build to the scientific inquiries we pursue.

The enduring significance of Newton’s Laws is a testament to their timeless relevance in science, technology, and education. As we journey further into the future, we carry with us the profound insights provided by these laws, reminding us of the boundless possibilities that lie ahead.

Call to Action

We encourage you to share this article with others who may be interested in the significance of Newton’s Laws. Additionally, if you have any questions or would like to share your insights on this topic, please leave a comment below. For those who wish to dive deeper into physics and scientific exploration, we recommend exploring the extensive resources available at academic institutions, physics organizations, and reputable educational websites.

References and External Links

We hope you’ve enjoyed this comprehensive exploration of the importance of Newton’s Laws of Motion. Your journey into the world of physics has only just begun, and we invite you to continue your quest for knowledge and understanding.

Sources

  1. Glenn Research Center
  2. Sparknote
  3. Library of Congress

what is the system’s potential energy when its kinetic energy is equal to 34e?

Solution

what is the system’s potential energy when its kinetic energy is equal to 34e?

The answer to the above question is (KA2)/8

video solution

Explanation

Law of Conservation of Energy

Law of conservation of energy states that energy can neither be created nor destroyed but can be transferred from one medium to another.

Additionally, the law states that in an isolated or closed system, the total amount of energy is always constant, although energy may be changed from one form to another.

Data

what is the system's potential energy when its kinetic energy is equal to 34e?
what is the system’s potential energy when its kinetic energy is equal to 34e?

The net energy in a system is

Net Energy (Enet) = Potential Energy (P.E) + Kinetic Energy (K.E)

Enet = P.E + K.E

We identify the net energy in a simple harmonic motion as

Enet = (1/2) Kx2

Hence, we can insert the above equation into Net Energy (Enet) = Potential Energy (P.E) + Kinetic Energy (K.E) to obtain

(1/2) Kx2 = P.E + K.E

and from our question, the kinetic energy, K.E = (3/4)E

Let’s assume that E = Net Energy = Enet = (1/2) Kx2

We can now rewrite the kinetic energy as

K.E = (3/4) x E = (3/4) x Enet = (3/4) x (1/2) Kx2 = (3/8) Kx2

Thus, the kinetic energy is K.E = (3/8) Kx2

You may also like to read:
Simple Harmonic Motion Formulae

Solution

we can now insert the above expression into (1/2) Kx2 = P.E + K.E to get

(1/2) Kx2 = P.E + (3/8) Kx2

It is now time to make the potential energy (P.E) subject of the formula

P.E = (1/2) Kx2 – (3/8) Kx2

The above equation can be simplified into

P.E = (1/2) Kx2 – (3/8) Kx2 = (1/2 – 3/8) Kx2 = ((4-3) /8 ) Kx2 = (1/8) Kx2

Therefore, the system’s potential energy is (1/8) Kx2

Drop a question in the comment section if you have a challenge with the question: “what is the system’s potential energy when its kinetic energy is equal to 34e?”

You may also like to read:

A Small Rock with Mass 0.20 kg is Released

a photon with an energy of 1.33

The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle.

what is the object’s velocity when its potential energy is 23e?

Sources:

Libretext

Newton’s Second Law Practice Problems

Newton’s Second Law Practice Problems

Here are Newton’s second law practice problems to help you master the calculations involving the second law of motion.

Newton’s Second Law Practice Problems

Problem 1

An unbalanced force of 20 Newtons acts on a 4.0-kilogram mass. What acceleration does it give it?

Newtons Second Law Practice Problems
Newton’s Second Law Practice Problems

Data

Force acting on the body, f = 20 N

Mass of the body, m = 4 kg

Unknown

Acceleration of the body, a = ?

Formula

The formula for Newton’s second law of motion is F = ma

and to find the acceleration, we need to make a subject of the formula

a = f/m

We will apply the above formula to solve the problem

Solution

Substitute your data into the formula

a = 20/4 = 5 ms-2 [ Because the unit of acceleration is ms-2 ]

Therefore, the acceleration of the body is 5 meters per second square.

Problem 2

A 900 grams stone is pushed along a tarmac by a horizontal force of 20 Newton. A frictional force of 8 newtons opposes the motion. What is the acceleration given to the stone?

Data

Mass of the stone, m = 900 g = (900/1000) kg = 0.9 kg

Horizontal force, fh = 20 N

Frictional force, fr = 8 N

Unknown

Unbalanced force, F = fh – fr = 20 – 8 = 12 N

The acceleration, a = ?

Formula

Force = mass x acceleration

Which can be written as

F = m x a

Solution

Now we can insert our data into the formula

The acceleration, a = F/m = 12/0.9 = 13.3 ms-2

Therefore, the acceleration of the stone is 13 ms-2

Problem 3

A ball of mass 0.3 kilograms moving at a velocity of 20 meters per second is suddenly hit by a force of 5 newtons for a time of 0.03 seconds. Find its new velocity of motion.

Data

Mass of the ball, m = 0.3 kg

Initial velocity, u = 20 m/s

Force, F = 5 N

Time, t = 0.03 s

Unknown

Final velocity, v = ?

Formula

Since F = ma

and a = (v-u)/t

We can see that

F = m(v-u)/t = (mv – mu)/t

We can make v subject of the formula from F = m(v-u)/t

F/m = (v-u)/t

and now

Ft/m = v – u

Therefore, the formula we need to apply is

v = (Ft/m) + u

Solution

It is now time for us to put our data into the formula

v = (5 x 0.03/0.3) + 20

Now, our final velocity, v = (0.15/0.3) + 20 = 0.5 + 20 = 20.5 m/s

Therefore, our final velocity is 20.5 meters per second

Problem 4

A body of mass 0.1 kilograms dropped from a height of 8 meters onto a hard floor and bounces back to a height of 2 meters. Calculate the change of momentum. If the body is in contact with the floor for 0.1 seconds, what is the force exerted on the body? ( Take gravitational acceleration, g = 10 ms-2).

Data

Mass of the body, m = 0.1 kg

Before hitting the floor, the height h1 = 8 m

After hitting the floor, the height of bouncing h2 = 2 m

Gravitational acceleration, g = 10 ms-2

Time, t = 0.1 s

Before hitting the floor, the initial velocity, u1 = 0

After hitting the floor, the final velocity after bouncing, v2 = 0

Unknown

Before hitting the floor, the final velocity, v1 = ?

After hitting the floor, the initial velocity after bouncing, u2 = ?

Formula

The first formula

v12 = u12 + 2gh1

The second formula

v22 = u22 – 2gh2

The final formula to apply is

F = (mv1 – (-mu2))/t = (mv1 + mu2)/t [For the body bouncing and moving away from the floor, the momentum is negative]

Solution

We can apply our data to the first formula

v12 = u12 +2gh1 = u12 +2gh1 = 0 + 2 x 10 x 8 = 160

Therefore

v12 = 160

and v1 = √160 = 12.7 m/s = 13 m/s

Applying our data to the second formula, we will have

v22 = u22 – 2gh2

0 = u22 – 2 x 10 x 2

u22 – 40 = 0

We can now collect like terms to obtain

u22 = 40

u2 = √40 = 6.325 m/s = 6 m/s

Applying the final formula, we can now find the force exerted on the body

F = (mv1 + mu2)/t

substituting our data and the rest of the values, we will get

Force, F = (mv1 + mu2)/t = m(v1 + u2)/t = 0.1(13 + 6)/0.1 = 19 N

Therefore, the force exerted on the body is 19 Newton

Problem 5: Newton’s Second Law Practice Problems

A body of mass 2 kilograms falls freely from rest, through a height of 50 meters, and comes to rest having penetrated 5.0 centimeters of sand.

a. Calculate the velocity with which the body hits the sand

b. The time taken in falling

c. The average force exerted by the sand in bringing the body to rest.

Data

Mass of the body, m = 2 kg

Initial height, h1 = 50 m

Final height, h2 = 5 cm = 5/100 m = 0.05 m

Gravitational acceleration, g = 10 ms-2

Initial velocity, u = 0

Unknown

v = ?

t = ?

retardation, a = ?

F = ?

Formula

The first formula is

v2 = u2 + 2gs

The second formula to apply is

v = u + gt

We will also apply v2 = u2 + 2as to find retardation

And then finally apply the formula for Newton’s second law of motion to solve the force exerted by the sand on the body

F = ma

Solutions

a. We can add our data into the equation v2 = u2 + 2gs

Hence,

v2 = 0 + 2 x 10 x 50 = 1000

v = √1000 = 31.623 m/s = 31 m/s

b. Substitute your data into the second formula

v = u + gt

we make t subject of the formula

v = gt [since u = 0]

Thus,

t = v/g = 31/10 = 3.1 s

c. We need to find the retardation by applying the formula v2 = u2 + 2as before we can go ahead to find the force

Insert our data into v2 = u2 + 2as

Note: After hitting the sand, our initial velocity = 31 m/s and the final velocity = 0

Hence,

0 = 312 + 2 x a x 0.05

We make a subject of the formula

a = -(961/0.1) = -9610 m/s2 [The negative sign indicates negative acceleration or retardation]

Therefore, a = 9610 m/s2

We can now comfortably apply our final formula that says Force (F) = Mass(m) x Acceleration (a)

Thus,

F = ma = 2 x 9610 = 19,220 N

Therefore, the force exerted by the sand in bringing the body to rest is 19,220 Newton

Problem 6: Newton’s Second Law Practice Problems

A towing van tows a car of mass 1,500 kilograms along a level road and accelerates at 0.8 meters per second square. What is the tension in the towing string?

If the towing string suddenly breaks when the car reaches a speed of 20 meters per second, calculate the distance covered by the car before it comes to rest on the application of a braking force of 2000 newtons.

Data

Mass of the car, m = 1,500 kg

Acceleration, a = 0.8 ms-2

Initial speed, u = 20 m/s

Final speed, v = 0 [Because the car finally came to rest, then its final velocity is zero]

Braking force = 2000 N

Unknown

The tension of the string, F = ?

Change in momentum = ?

Time taken, t = ?

Deceleration, aD = ?

Distance covered, s = ?

Formula

To find the tension of the string, we will apply

F = ma

For change in momentum, the formula will be

Change in momentum = mu – mv

The to find time taken is

t = change in momentum / tension on the string (F)

Finding deceleration, we will apply

a = (v – u) / t

And finally, the last formula to find the distance covered, we use

s = (v2 – u2) / 2a

Solution

Let us find the tension on the string by substituting our data into F = ma

F = 1,500 x 0.8 = 1,200 N

The change in momentum = mv – mu = m(v – u) = 1,500 (20 – 0) = 1,500 x 20 = 30,000 kgms-1

To calculate the time taken, we will insert our data into the formula

Time taken, t = change in momentum / tension on the string = 30,000 / 1,200 = 25 seconds

The deceleration will be

aD = (v – u) / t = (0 – 20) / 25 = – 0.8 ms-2

Now, we can go ahead to calculate the distance covered

s = (v2 – u2) / 2aD = (02 – 202) / 2 x (-0.8) = -400/-1.6 = 250 m

Therefore, the distance covered is 250 meters

Therefore, I hope these Newton’s second law practice problems have helped you understand how to solve problems related to the second law of motion.

You can share your biggest challenge in Newton’s second law practice problems so that i can help you solve them.

You may also like to read:

Newton’s Second Law of Motion Examples

15 Newton’s Laws of Motion Project Ideas

Sources:

Britannica

Speed, Velocity, and Acceleration Problems

What is Speed?

Definition of Speed: Speed is the rate of change of distance with time. Speed is different from velocity because it’s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Additionally, you need to know that speed is a scalar quantity and we can write its symbol as S. The formula for calculating the speed of an object is:

Speed, S = Distance (d) / Time (t)

Thus, s = d/t

Note: In most cases, we also use S as a symbol for distance.

The S.I unit for speed is meter per second (m/s) or ms-1

Non-Uniform or Average Speed: This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

The formula for calculating non-uniform speed is

Average speed = Total distance covered by the object / Total time taken

Actual speed: This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

You may also like to read:

How to Find Displacement in Physics

also How to Find a Position in Physics

and How to Calculate Bearing in Physics

What is Velocity?

Definition of Velocity: Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms-1). We use V as a symbol for velocity.

Note: We often use U to indicate initial speed, and V to indicate final speed.

The formula for calculating Velocity (V) = displacement (S) / time (t)

The difference between velocity and speed is the presence of displacement and distance respectively. Because displacement is a measure of separation in a specified direction, while distance is not in a specified direction. Velocity is a vector quantity.

Uniform Velocity

Definition of uniform velocity: The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

The formula for uniform velocity = Total displacement / Total time taken

What is Acceleration?

Definition of Acceleration: Acceleration is the rate of change of velocity with time. Acceleration is measured in meters per second square (ms-2). The symbol for acceleration is a. Acceleration is also a vector quantity.

The formula for acceleration, a = change in velocity (v)/time taken (t)

Thus, a = v/t

We can also write acceleration as

a = change in velocity/time = ΔV/Δt = (v – u)/t

[where v = final velocity, u = initial velocity, and t = time taken]

Uniform Acceleration

In the case of uniform acceleration, the rate of change of velocity with time is constant.

Average velocity of the object = (Initial velocity + final velocity)/2

or

Average velocity = (v + u)/2

What is Retardation?

Retardation is the decreasing rate of change in velocity moved, covered, or traveled by an object.

The formula for calculating retardation is

Retardation = Change in a decrease in velocity/time taken

Equations of Motion

You can also apply the following equations of motion to calculate the speed, velocity, and acceleration of the body:

  1. v = u + at
  2. s = [ (v + u)/2 ]t
  3. v2 = u2 + 2as
  4. s = ut + (1/2)at2

Where v = final velocity, u = initial velocity, t = time, a = acceleration, and s = distance

You may also like to read:

How to Calculate Maximum height

What are Prefixes in Physics

How to Calculate Dimension in Physics

Solved Problems of Speed, Velocity, and Acceleration

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

Problem 1

A train moves at a speed of 54km/h for a one-quarter minute. Find the distance travelled by train.

Speed, Velocity, and Acceleration Problems
Speed, Velocity, and Acceleration Problems

Solution

Data

From the question above

Speed = 54 km/h = (54 x 1000)/(60 x 60) = 54,000/3,600 = 15 m/s

Time = one quarter minute = 1/4 minute = (1/4) x 60 = 15 seconds

Since we have

speed = distance/time

After cross-multiplication, we will now have

Distance = speed x time

We can now insert our data into the above expression

Distance = 15 m/s x 15 s = 225 m

Therefore, the distance travelled by train is 225 meters.

Problem 2

A car travelled a distance of 5km in 50 seconds. Find the speed in meters per second.

Speed, Velocity, and Acceleration Problems

Solution

Data:

Distance = 5km = 5 x 1000m = 5,000m

Time = 50 seconds

speed = ?

and the formula for speed

speed = distance/time = 5000/50 = 100m/s

Problem 3

A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration.

Solution

Data:

From the question above

Initial velocity, u = 0 (because the motorcycle starts from rest)

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Time taken, t = 15 seconds

Therefore, we can now apply the formula that says

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms-2

Therefore, the acceleration of the motorcycle is 2ms-2

Problem 4

A bus covers 50 kilometres in 1 hour. What is it is the average speed?

Solution

Data

Total distance covered = 50 km = 50 x 1000m = 50,000m

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Average speed = Total distance covered by the object / Total time taken

Therefore, we can now calculate the average speed of the bus by substituting our data into the above formula

Hence,

Average speed = 50,000/3,600 = 13.9 m/s

Therefore, the average speed of the bus is 13.9m/s or approximately 14 meters per second.

Problem 5

A car travels 80 km in 1 hour and then another 20 km in the next hour. Find the average velocity of the car.

Speed, Velocity, and Acceleration Problems

Solution

Data:

Initial displacement of the car = 80km

Final displacement of the car = 20km

The total displacement of the car = initial displacement of the car + final displacement of the car

Therefore,

The total displacement of the car = 80km + 20km = 100km

Also,

The time for 80km is 1hr

And the time for 20km is 1hr

Total time taken = The time for 80km (1hr) + The time for 20km (1hr)

Hence

Total time taken = 1hr + 1hr = 2hrs

Now, to calculate the average velocity of the car, we apply the formula that says

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

We can further convert the above answer into meters per second

Average velocity = 50km/h = (50 x 1000m)/(1 x 60 x 60s) = 50,000/3,600 = 13.9m/s or 14ms-1

Therefore, the average velocity of the car is 14ms-1

You may also like to read:

How to Conduct Physics Practical

and How to Calculate Velocity Ratio of an Inclined Plane

Problem 6

A body falls from the top of a tower 100 meters high and hits the ground in 5 seconds. Find its acceleration.

Solution

Data

Displacement = 100m

Time = 5 seconds

and we can apply the formula for acceleration that says

acceleration, a = Displacement/time2 = 100/52 = 100/25 = 4ms-2

Therefore the acceleration due to the gravity of the body is 4ms-2

Note: The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms-2), even though there may be a slight difference due to the mass and altitude of the body.

Problem 7

An object is thrown vertically upward at an initial velocity of 10ms-1 and reaches a maximum height of 50 meters. Find its initial upward acceleration.

Solution

Data

Initial velocity, u = 10ms-1

Final velocity, v = 0

maximum height = displacement = 50m

Initial upward acceleration, a =?

When we apply the formula that says a = (v2 – u2)/2d we will have

a = (0 – 102)/(2 x 50) = -100/100 = -1 ms-2

Hence, since our acceleration is negative, we can now say that we are dealing with retardation or deceleration.

Therefore, the retardation is -1ms-2

Note: Retardation is the negative of acceleration, thus it is written in negative form.

Problem 8

A car is traveling at a velocity of 8ms-1 and experiences an acceleration of 5ms-2. How far does it travel in 4 seconds?

Solution

Data:

Initial velocity, u = 8ms-1

acceleration, a = 5ms-2

Distance, s =?

time, t = 4s

We can apply the formula that says

s = ut + (1/2)at2

Thus

s = 8 x 4 + (1/2) x 5 x 42

s = 32 + (1/2) x 80 = 32 + 40 = 72m

Therefore, the distance covered by the car in 4s is 72 meters.

Problem 9

A body is traveling at a velocity of 10m/s and experiences a deceleration of 5ms-2. How long does it take the body to come to a complete stop?

Solution

Data

Initial velocity, u = 10m/s

Final velocity, v = 0

acceleration , a = retardation = -5ms-2

time, t = ?

We already know that acceleration, a = change in velocity/time

This implies that

Time, t = change in velocity/acceleration

Thus

t = (v – u)/t = (0-10)/-5 = -10/-5 = 2s

Therefore, the time it takes the car to stop is 2 seconds.

Problem 10

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

Solution

Data

Change in velocity, v =10 m/s

mass of the object, m = 10kg

time, t = 5s

We can apply newton’s second law of motion which says f = ma

and since a = change in velocity/time

we will have an acceleration equal to

a = 10/5 = 2ms-2

Therefore, to find the force, we can now say

f = ma = 10 x 2 = 20N

Therefore, the force that can help us to change the velocity by 10 m/s in 5 seconds is 20-Newton.

Drop a comment if there is anything you don’t understand about speed, velocity, and acceleration Problems.

You may also like to read:

How to Calculate Cost of Electricity Per kwh

also How to Calculate the Efficiency of a Transformer

and How to Calculate Escape Velocity of a Satellite

Check our Website:

Wokminer – Apply for jobs

How to Calculate the Relative Density of a Liquid

What is Relative Density?

Definition and Concept of Relative Density: Relative density, also known as specific gravity, is a dimensionless quantity that describes the ratio of the density of a liquid to the density of a reference substance. In this article, I will provide you with a comprehensive explanation of how to calculate the relative density of a liquid, its definition, and its formula. I will also provide you with a video explanation to broaden your understanding of the topic.

It is a measure of how heavy a liquid is compared to an equal volume of a chosen reference material. Watch the video below for more clarification.

This property, often referred to as specific gravity, provides insights into the density of a liquid compared to a reference substance, aiding in quality control, material identification, and even environmental monitoring.

Relative Density of Liquid Formula

How to Calculate the Relative Density of a Liquid
How to Calculate the Relative Density of a Liquid

The three relative density formula are as follows:

  1. Relative Density (RD) of Liquid = UL / UW
  2. RD = VL / VW
  3. Relative of liquid (RD) = AL / AW

Where:

RD = Relative density

UL = Upthrust of object in liquid

VL = Volume of liquid displaced by object

AL = Apparent loss of weight of a solid in liquid

UW = Upthrust of object in water

VW = Volume of water displaced by object

AW = Apparent loss of weight of same solid in water

We applied the relative density formula above to solve example 5 and 6

How to Calculate Relative Density

How to Calculate the Relative Density of a Liquid

Relative Density of Solids vs Relative Density of Liquids

Here is an expanded table with materials and their relative densities (liquid and solids):

MaterialStateRelative Density (g/cm³)
IronSolid7.87
LeadSolid11.34
GoldSolid19.32
CopperSolid8.96
AluminumSolid2.70
SilverSolid10.49
DiamondSolid3.52 – 3.53
IceSolid0.92
GraniteSolid2.63 – 2.75
Wood (Oak)Solid0.6 – 0.9
WaterLiquid1.00
Olive OilLiquid0.91
MercuryLiquid13.59
EthanolLiquid0.789
MilkLiquid1.02 – 1.05
HoneyLiquid1.36 – 1.45
Vegetable OilLiquid0.91 – 0.93
GlycerolLiquid1.26

You need to keep in mind that these values are approximate and can vary depending on various factors.

Explanation of How to Calculate the Relative Density of a Liquid

Here I will explain two steps you can apply so that you will understand how to calculate the relative density of a liquid

Step 1

I will use an empty dry bottle to explain in simple terms, how to calculate the relative density of liquids. Now, let us get started:

  1. Get a dry empty bottle, record it is mass and name it M1
  2. Add liquid (for example oil) to the bottle and name it M2. Now, mass of the liquid is added to the mass of the mass of the empty bottle and it is M2.
  3. The next step is to remove the liquid from the bottle.
  4. Wash and dry the bottle again
  5. Add water to the bottle, and record their total mass together. Tag this mass M3

To Find only the mass of the liquid without the bottle, subtract the Mass of the empty bottle (M1) from the Mass of the bottle and the liquid (M2) which is

Mass of liquid only = M2 – M1

Additionally, you need to apply the following methods to find the Mass of an equal volume of water:

subtract the Mass of the empty bottle (M1) from the Mass of the bottle and that of the water together (M3) which is:

Mass of equal volume of water = M3 – M1

Therefore, we can now say that the formula for how calculating the relative density of a liquid is

Relative density of liquid = Mass of the liquid / Mass of equal volume of water

which is

Relative density of liquid = (M2 – M1) / (M3 – M1)

You may also like to read:

Derivation of Bernoulli’s Equation

Step 2

In this step, I will apply Archimedes’ principle to explain how to calculate the relative density of liquid:

  1. Attach a metal object to spring balance and suspend it in air. Record the mass of the object in air as M1.
  2. Immerse the metal object (that is already attached to the spring balance) in liquid, and record the apparent mass of liquid as M2
  3. Clean and dry the metal object
  4. Immerse the object in the water and record the apparent mass of water as M3
  5. We will now find the weight of the liquid that is being displaced by the metal object, which is the upthrust in liquid.
  6. Now, the upthrust in liquids is calculated by subtracting mass of the object in air as M1 from apparent mass of liquid as M2. This implies that Upthrust in liquids = (M2M1)
  7. Upthrust in water is the same thing as the weight of the water displaced by the object
  8. To find upthrust in water, we subtract apparent mass of water (M3) from mass of the object in air (M1). Therefore, Upthrust in water = M1M3
  9. Hence, to calculate Relative density, we say that

Relative density = [ Upthrust in liquids = M2M1 ] / [Upthrust in water = M1M3]

Relative density = (M2M1) / (M1M3)

Examples of How to Calculate the Relative Density of Liquids

Remember to follow the following steps in order to solve your examples at ease:

  • Underline all the figures you see in the question
  • Read the question, and extract your data
  • Underline the statement where you see the question and write the formula you need to apply
  • Check all the elements in the formula and assign their figures to them
  • Solve the problem by making the element in question the subject of the formula
  • Don’t forget to always include the s.i unit in your final answer

Example 1

A body weighs 0.30N in air, 0.25N when fully immersed in water, and 0.27N when fully immersed in a liquid. Calculate (a) it is loss of weight in water, (b) the relative density, and (c) The relative density of the liquid

Solution:

(a) To calculate the loss of weight in water, we subtract the weight of the body when fully immersed in water (0.25N) from the weight of the body in the air (0.30N).

This implies that

Loss of weight in air = 0.30 – 0.25 = 0.05N

Therefore,

Loss of weight in air = Upthrust in water = Weight of equal volume of water displaced by the body = 0.05N

(b) Relative density of the body = Weight of the body in the air (0.30) / Weight of equal volume in water (0.05)

Now, we can see that

Relative density of the body = 0.30 / 0.05 = 6

(c) Relative density of liquid = Upthrust in liquid (0.30 – 0.27) / Upthrust in water (0.30 – 0.25)

This is because

Upthrust in liquid = mass of the body in air – mass of the body in liquid

and Upthrust in water = mass of the body in air – mass of the body in water

Thus,

Relative Density of the liquid = (0.30 – 0.27) / (0.30 – 0.25)

Therefore, the Relative density of liquid = 0.03 / 0.05 = 0.6

Now we can see that the Relative density of the liquid is 0.6

Example 2

A relative density bottle weighs 20 grams when empty, 80 grams when filled with water, and 100 grams when filled with a liquid. Find the relative density of the liquid.

Solution:

Data

Weight of the bottle in the air, W1 = 20 grams

Also, the weight of the bottle with water, W2 = 80 grams

Additionally, the weight of the bottle with liquid, W3 = 100 grams

Applying the formula

Relative density = (W3 – W1) / (W2 – W1)

Thus, we have

Relative density = (100 – 20) / (80 – 20) = 80 / 60 = 1.33

Example 3

A density bottle has a mass of 0.05 kilogram when empty, a mass of 0.30 when some quantity of steel ball bearings is added to it, and a mass of 0.35 kilogram when the remainder of the bottle is filled with water. if the density of the bottle weighs 0.2 kilograms when filled with water. Calculate the relative density of the steel ball bearing.

Solution:

Data

Mass of an empty bottle, M1 = 0.05 kg

Also, the Mass of the bottle with steel ball bearings, M2 = 0.30 kg

Mass of the bottle, steel ball bearings, and water, M3 = 0.35

we can also see that the Mass of the bottle with water, M4 = 0.2 kg

The Relative density of the steel (R.D) = ?

Now, let us apply the formula relative density (r.d) to solve the above problem

Mass of the steel ball bearing M5 = M2 – M1 = 0.30 – 0.05 = 0.25 kg

We can also see that

Mass of water and the bottle M6 = M4 – M1 = 0.2 – 0.05 = 0.15 kg

Additionally, mass of water, the bottle, and the steel ball bearing M7 = M3 – M2 = 0.35 – 0.30 = 0.05 kg

We finally calculate the mass of water having the same volume as the ball bearing M8 = M6 – M7 = 0.15 – 0.05 = 0.10

Therefore to find the relative density of steel ball bearings, we say that

Relative density (R.d) = M5 / M8 = 0.25 / 0.10 = 2.5

Example 4

The mass of a bottle is 18 grams when it is empty, 44.0 grams when it is filled with water, and 39.84 gram when it is full of a second liquid. Calculate the density of the liquid.

Solution:

Data:

Mass of an Empty bottle M1 = 18.0 g

Also,

Mass of the bottle and that of water M2 = 44.0 g

Therefore, we have mass of water, M3 = M2 – M1 = 44.0 – 18.0 = 26 g

Also, the mass of the bottle and the liquid M4 = 39.84 g

Now, mass of liquid, M5 = M4 – M1 = 39.84-18.00 = 21.84 g

The relative density of the liquid = mass of liquid / mass of an equal volume of water = M5 / M3

R.d = 21.84/26 = 0.84

The density of the liquid = 0.84 g/cm3

Example 5

How to Calculate the Relative Density of a Liquid
How to Calculate the Relative Density of a Liquid

Example 6

How to Calculate the Relative Density of a Liquid
How to Calculate the Relative Density of a Liquid

Note:

Relative density is simply comparing the density of substances (solid, liquid, or gaseous) to that of water (1 g/cm3). It does not have a unit and may be measured with a hydrometer. We use the symbol R.D to indicate relative density, and it’s a scalar quantity

Method 2: How to Calculate the Relative Density of a Liquid

Relative density, often referred to as specific gravity, is the ratio of the density of a substance to the density of a reference substance, usually water. It provides information about the density of a liquid in comparison to water. To calculate the relative density of a liquid, follow these steps:

  1. Gather the Necessary Information: Before you begin, ensure you have the density of the liquid you want to measure and the density of water at a specific temperature. These values are typically provided in kilograms per cubic meter (kg/m³) or grams per cubic centimeter (g/cm³).
  2. Identify the Formula: The formula to calculate relative density is: Relative Density = (Density of Liquid) / (Density of Water)​
  3. Plug in the Values: Substitute the known values into the formula. Make sure the units are consistent to get an accurate result.
  4. Perform the Calculation: Use a calculator to divide the density of the liquid by the density of water. The result will be a dimensionless quantity, as both densities have the same units.
  5. Interpret the Result: The calculated relative density will provide you with valuable information about the liquid’s density compared to water. If the relative density is less than 1, the liquid is less dense than water, and if it’s greater than 1, the liquid is denser.

Method 3: Measurement and Calculation of Relative Density

A. Basic Formula: Relative Density = Density of Liquid / Density of Reference Substance The relative density is calculated by dividing the density of the liquid under consideration by the density of the chosen reference substance.

B. Choice of Reference Substance The selection of an appropriate reference substance is critical. Common choices include water, air, or a specific standardized solution, depending on the nature of the liquid being studied.

C. Experimental Techniques for Determining Relative Density Several techniques are employed to determine the relative density of liquids:

  1. Hydrometer Method: This involves using a hydrometer, a device that measures the buoyancy of a liquid. The depth to which the hydrometer sinks is related to the relative density of the liquid.
  2. Pycnometer Method: A pycnometer is a container specifically designed to hold a known volume of liquid. By measuring the mass of the liquid it contains, and comparing it to an empty pycnometer, the relative density can be calculated.
  3. Buoyancy Method: Archimedes’ principle is applied in this method, where the buoyant force exerted on an immersed object is used to determine the relative density of the liquid.

Importance of Relative Density in Various Applications

The concept of relative density finds application across a multitude of fields. From ensuring the quality of beverages and identifying unknown liquids to aiding engineers in selecting appropriate fluids for various processes, relative density is a cornerstone in understanding the behaviour and characteristics of liquids.

Relation to Density and Specific Gravity

While density measures the mass of a substance per unit volume, specific gravity is a dimensionless quantity that compares the density of a substance to the density of water at a specified temperature. Relative density and specific gravity are closely related, with the latter often being used interchangeably to express the former.

Factors Affecting Relative Density

A. Temperature Effects Temperature variations can influence the relative density of liquids due to changes in volume and density. Accurate measurements must account for temperature fluctuations.

B. Pressure Effects Pressure changes can also impact relative density, particularly in gases and compressible liquids. Corrections are necessary when working under non-standard pressure conditions.

C. Impurities and Composition of the Liquid Impurities and variations in composition can affect the relative density, making it important to consider the purity of the liquid under examination.

Significance and Applications of Relative Density

A. Quality Control in Industries Relative density is employed in industries such as food and beverages, where it aids in monitoring product consistency and detecting adulteration.

B. Identification of Liquids In forensic analysis and material identification, relative density can be a valuable tool for determining the nature of unknown liquids.

C. Selection of Fluids in Engineering Engineers rely on relative density to choose the appropriate liquids for hydraulic systems, cooling processes, and other engineering applications.

D. Determination of Concentration in Solutions Relative density assists in determining the concentration of solutes in solutions, such as in the pharmaceutical and chemical industries.

E. Monitoring Changes in Mixtures and Blends Relative density is used to track changes in mixtures and blends, aiding in quality assurance during production processes.

Comparison with Other Density-related Concepts

A. Absolute Density While relative density compares a liquid’s density to that of a reference substance, absolute density stands alone as the density of the substance itself.

B. Specific Gravity Specific gravity is essentially synonymous with relative density, often used in industrial contexts and practical applications.

C. Bulk Density Bulk density considers the density of a substance in bulk, accounting for voids and air spaces between particles.

Importance in Daily Life and Industry

A. Beverages and Food Industry Relative density ensures the consistency and quality of beverages and food products, preventing consumer health risks and ensuring accurate labeling.

B. Petroleum and Petrochemical Industry In the oil and gas sector, relative density is pivotal in classifying and evaluating petroleum products.

C. Pharmaceutical and Medical Applications Pharmaceutical companies rely on relative density for drug formulation and dosage determination.

D. Environmental Monitoring Relative density aids environmental scientists in analyzing pollutants, tracking water quality, and assessing ecosystem health.

Challenges and Limitations in Measuring Relative Density

A. Accuracy and Precision Accurate measurement of relative density requires advanced equipment and careful techniques to minimize errors.

B. Impact of Viscosity Viscous liquids may pose challenges, as their flow characteristics can affect measurement accuracy.

C. Compatibility with Different Reference Substances The choice of reference substance must be appropriate for the liquid under investigation.

Examples and Case Studies

A. Determination of Alcohol Content in Beverages Relative density is utilized to determine the alcohol content in alcoholic beverages, contributing to regulatory compliance and consumer safety.

B. Fuel Quality Testing in Automobile Industry The automotive sector employs relative density to assess fuel quality and efficiency.

C. Identification of Unknown Liquids in Forensics Forensic scientists use relative density to identify liquids found at crime scenes, aiding in criminal investigations.

Future Trends and Developments in Relative Density Measurement

A. Advances in Measurement Techniques Continued advancements in technology will likely lead to more precise and efficient methods for measuring relative density.

B. Integration with Automation and Technology Automation and digitalization will streamline data collection and analysis, enhancing accuracy and reducing human error.

FAQs about Relative Density of a Liquid

Is relative density the same as density?

No, relative density and density are not the same. Density is an absolute measure of how much mass is contained in a given volume, while relative density compares the density of a substance to the density of another substance, typically water.

Can relative density be less than 1?

Yes, a substance’s relative density can be less than 1. This means that the substance is less dense than water and will float on its surface.

How is relative density measured?

Relative density is measured using specialized instruments called hydrometers or densimeters. These devices are designed to measure the density of a liquid and compare it to the density of water.

Why is water used as the reference substance for relative density?

Water is commonly used as the reference substance for relative density because of its widespread availability and consistent properties. Additionally, water has a well-defined density of 1.0 g/cm³ at a specific temperature, making it a convenient reference point.

What is the importance of relative density in the beverage industry?

In the beverage industry, relative density is crucial for quality control and consistency. It helps ensure that beverages are properly concentrated and mixed, resulting in the desired taste and experience for consumers.

How does relative density affect sea life?

Relative density plays a role in the movement of marine organisms. Some species have specific adaptations that allow them to control their buoyancy by adjusting their internal gas volume, helping them rise or sink in the water.

Reference

  1. First
  2. Smith, J. D. (2008). Introduction to chemical engineering thermodynamics. McGraw-Hill.
  3. Perry, R. H., Green, D. W., & Maloney, J. O. (1984). Perry’s Chemical Engineers’ Handbook (6th ed.). McGraw-Hill.
  4. ASTM International. (2017). ASTM E100-17: Standard Specification for ASTM Hydrometers. West Conshohocken, PA.
  5. Callister, W. D., & Rethwisch, D. G. (2017). Materials Science and Engineering: An Introduction (10th ed.). John Wiley & Sons.

Derivation of Bernoulli’s Equation

What is Bernoulli’s Equation?

Derivation of Bernoulli’s Equation: Bernoulli’s theorem states that the sum of pressure, potential energy per unit volume, and kinetic energy per unit volume of an ideal fluid that is flowing in a steady state is constant at each point along a streamlined path in a pipe.

Additionally, Bernoulli’s equation is a fundamental principle in fluid dynamics that explains how the pressure, velocity, and elevation of a flowing fluid are related. It helps us understand how fluids, like water or air, move in pipes or over surfaces.

Derivation of Bernoulli's Equation
Derivation of Bernoulli’s Equation

Imagine you are holding a hose and water is flowing out. Bernoulli’s equation tells us that as the water moves faster (higher velocity), its pressure decreases. This is why you can feel the hose pushing back when you cover part of the opening – the pressure increases, slowing down the water.

Additionally, if you lift the hose, the water flows slower at the top, and according to Bernoulli’s equation, its pressure increases compared to the bottom where it’s flowing faster.

This principle is used in many real-life situations. For example, airplanes stay in the air because the shape of their wings makes air move faster above them, reducing pressure and creating lift. Formula 1 cars use it to stay glued to the track by designing their shapes to control airflow. In essence, Bernoulli’s equation helps us understand how fluid speed and pressure changes are interconnected.

Bernoulli’s Equation Formula

At the heart of fluid dynamics lies the cornerstone known as Bernoulli’s equation. This equation, derived from the principle of conservation of energy, describes the behaviour of a fluid as it flows through different sections of a pipeline or conduit. In its simplest form, the equation can be expressed as follows:

P + ½ρv² + ρgh = constant

Where:

  • P represents the pressure exerted by the fluid,
  • ρ denotes the fluid density,
  • v symbolizes the fluid velocity,
  • g stands for the acceleration due to gravity, and
  • h represents the height above a reference point

Video Explanation on Derivation Bernoulli’s Equation

Part One: Derivation of Bernoulli’s Equation
Part Two: Derivation of Bernoulli’s Equation

The Components of Bernoulli’s Equation

Breaking down Bernoulli’s equation reveals its underlying components, each contributing to the overall understanding of fluid dynamics:

  1. Pressure Term (P): The pressure exerted by the fluid on its surroundings plays a crucial role in determining the fluid’s behavior. As the fluid’s speed changes, its pressure also varies, influencing the dynamics of the system.
  2. Kinetic Energy Term (½ρv²): This term represents the kinetic energy possessed by the fluid due to its velocity. It highlights the connection between the fluid’s speed and its energy state.
  3. Potential Energy Term (ρgh): The potential energy term takes into account the fluid’s height above a reference point. It demonstrates the influence of gravitational potential energy on the fluid’s overall energy.

By comprehending these components, we can unlock the equation’s true potential and apply it to a myriad of practical scenarios.

Derivation of Bernoulli’s Equation:

At the heart of fluid dynamics lies Bernoulli’s equation, which describes the relationship between pressure, velocity, and elevation in a fluid flowing along a streamline. This equation is a manifestation of the conservation of energy principle applied to fluid flow. By considering various factors, we can derive Bernoulli’s equation step by step.

Conservation of Energy:

The derivation of Bernoulli’s equation starts with the concept of conservation of energy. This fundamental principle asserts that the total mechanical energy of a fluid particle remains constant along its path, neglecting external forces like friction.

Kinetic Energy Term:

The kinetic energy term in Bernoulli’s equation accounts for the energy associated with the fluid’s velocity. As a fluid particle moves, its kinetic energy changes, influencing the overall energy balance.

Potential Energy Term:

The potential energy term introduces the elevation of the fluid particle in a gravitational field. Changes in potential energy contribute to variations in the total energy of the fluid.

Pressure Energy Term:

The pressure energy term represents the force exerted by the fluid’s pressure. It plays a crucial role in the derivation as pressure differences contribute significantly to fluid flow behavior.

Simplifying Assumptions:

To facilitate the derivation process, certain assumptions are made. These include steady flow conditions, incompressibility of the fluid, and negligible viscous effects.

Applying Euler’s Equation:

Euler’s equation, derived from Newton’s second law, is a fundamental equation in fluid dynamics. It forms the basis for deriving Bernoulli’s equation by considering pressure, velocity, and acceleration terms.

Integrating Euler’s Equation:

Integrating Euler’s equation along a streamline leads to the development of Bernoulli’s equation. This integration combines the kinetic energy, potential energy, and pressure energy terms, offering a comprehensive insight into fluid flow behavior.

Applications of Bernoulli’s Equation:

The derivation of Bernoulli’s equation opens the door to a multitude of practical applications in various fields. From aviation to hydraulics, this equation plays a pivotal role in understanding and predicting fluid behavior.

Aerodynamics and Flight:

In aviation, Bernoulli’s equation elucidates the principles of lift generation. The faster-moving air above an aircraft’s wing corresponds to lower pressure, creating an upward force that allows flight.

Blood Flow and Physiology:

The circulatory system’s functioning can be explained using Bernoulli’s equation. Blood flow through arteries and veins involves pressure differences, which influence the circulation of blood in the body.

Hydraulic Engineering:

Hydraulic systems heavily rely on Bernoulli’s equation to design pipelines, water supply networks, and dams. This equation aids in optimizing fluid transport and distribution.

Meteorology and Weather Patterns:

Bernoulli’s equation contributes to understanding weather phenomena such as wind patterns and air pressure variations. It explains how differences in atmospheric pressure drive air movements on a global scale.

Sports and Recreational Activities:

Sports like baseball and golf exploit the principles of Bernoulli’s equation. The phenomenon of a curveball in baseball and the lift of a golf ball are both influenced by fluid dynamics.

Bernoulli’s Equation in Action: Practical Examples

To solidify our understanding, let’s delve into practical scenarios where Bernoulli’s equation takes center stage:

1. Airplane Lift and Flight

Picture yourself seated in an airplane, soaring through the sky. Have you ever wondered how this massive machine defies gravity? The answer lies in Bernoulli’s equation. As air flows over the curved shape of the wing, it accelerates, leading to a decrease in pressure above the wing and a higher pressure below. This pressure difference creates lift, allowing the aircraft to achieve and maintain flight.

2. The Marvel of Straw Sippers

Ever marveled at the physics behind a straw sipper? When you sip a beverage through a straw, you create a low-pressure area within the straw. Bernoulli’s equation explains this phenomenon: the increased fluid velocity within the narrow straw results in lower pressure, causing the liquid to rise into the straw and into your mouth.

3. Squirt Guns and Fluid Dynamics

Engage in some summertime fun with a squirt gun, and you’re also experimenting with Bernoulli’s equation. As you squeeze the trigger, water is forced through a narrow nozzle. The increased velocity of the water results in a drop in pressure, allowing atmospheric pressure to push water into the gun, ready for your next squirt.

FAQs about Bernoulli’s Equation:

What is the fundamental concept behind Bernoulli’s equation?

Bernoulli’s equation is based on the conservation of energy principle, stating that the total mechanical energy of a fluid particle remains constant along its path in the absence of external forces like friction.

How does Bernoulli’s equation relate to fluid flow?

Bernoulli’s equation establishes a relationship between pressure, velocity, and elevation in a fluid flowing along a streamline. It helps predict how these factors influence fluid behavior.

Why is Bernoulli’s equation important in aviation?

In aviation, Bernoulli’s equation explains lift generation. Faster air above an aircraft wing results in lower pressure, creating an upward force that enables flight.

Can Bernoulli’s equation be applied to non-ideal fluids?

Bernoulli’s equation assumes incompressibility and neglects viscous effects, making it suitable for ideal fluids. For non-ideal fluids, modifications and additional terms are necessary.

How does Bernoulli’s equation impact hydraulic system design?

Hydraulic engineering relies on Bernoulli’s equation to optimize the design of pipelines, water distribution networks, and hydraulic machinery for efficient fluid transport.

Is Bernoulli’s equation applicable to gases as well as liquids?

Yes, Bernoulli’s equation applies to both gases and liquids, as long as the assumptions of incompressibility and negligible viscosity hold true for the specific fluid.

You may also like to read:

A Sample of Physics Lesson Plan

Reference:

Common Applications of Bernoulli’s Principle

Physics Practical: How to do Physics Practical

 

What is Physics Practical?

Physics practical refers to the hands-on experimentation and observation of physical phenomena to better understand and verify the principles and theories of physics. It involves conducting experiments, measurements, and data collection in a controlled environment, often using specialized equipment and instruments.

Additionally, Physics practicals are very important because they help students understand and apply theoretical concepts in real-world situations. They enable hands-on learning, allowing students to directly interact with materials and equipment, fostering a deeper comprehension of physics principles.

Through experiments, students can verify theoretical predictions, develop problem-solving skills, and learn how to gather and analyze data. Practical work also cultivates critical thinking by encouraging students to make observations, draw conclusions, and refine their hypotheses.

Moreover, physics practicals nurture essential laboratory skills, such as measurement accuracy and safety protocols, which are valuable in scientific research and various careers. Overall, practicals bridge the gap between theory and practice, enhancing students’ overall understanding and appreciation of the fundamental laws that govern the natural world.

physics practical
physics practicals

You may also like to read:

How to Calculate and Find Capacitive Reactance

What is the difference between Theory and Experiment

When we talk about theory, we are simply referring to a statement of fact due to the availability of experimental evidence to support it. A theory may be disapproved if there is no experimental evidence to support it.

Whenever we develop a theory, we must perform an experiment to ascertain its existence. 

By performing an experiment, we will be able to monitor and accumulate data that can support the theory we earlier came up with. 

The theory is an idea, while an experiment is practical that supports the existence of the idea. 

How to Report a Physics Experiment  

Here are the procedures one needs to follow to report a physics experiment:

  1. You need to come up with a title for the Experiment
  2. The next step is to write the aim of your experiment. 
  3. You now mention the names of the apparatuses you use for the experiment. 
  4. Draw a diagram of how the apparatuses were set up and label your diagram 
  5.  Draw a table for recording your data 
  6. Fill in the table with the results you obtain from the series of individual experiments. 
  7. State the theory of the experiment to back up your reason why the procedures and the readings are mandatory. 
  8. Apply the equation you need to make certain calculations to come up with your readings 
  9. Draw your graph 
  10. Find the gradient or slope of the graph 
  11. In some cases, you may also need to apply the result from your slope to make calculations. The formula will be given to you. 
  12. State the precautions you observed while carrying out the practical to get an accurate reading. 
  13. You also need to avoid common errors while carrying out a physics experiment. 
  14. Additionally, you need to wear a lab coat before entering the Physics Laboratory. You need to focus on your practical skills to avoid making unnecessary mistakes. 

How to Plot and Calculate a Graph 

To plot a beautiful graph, you will need to maintain calmness and concentrate on the practical. By plotting a graph, you will understand the accuracy of your data. 

The aim of plotting a graph is to show the relationship between variable quantities during physics practicals.

After plotting a graph, you will need to find the following:

  • Gradient or slope 
  • The intercept on the x-axis or y-axis.
  • Apply any of the formulas they give you to make your deductions. 

The items you will need to plot a graph include:

  • A set of mathematical set
  • A sharp pencil 
  • A transparent ruler 
  • A razor or sharpener 
  • A neat graph book 
  • A broomstick for drawing curves 
  • A biro 

What types of graphs we can plot?

To understand how to conduct physics practicals, we must first know three types of graphs and they are:

  1. Straight line graphs
  2. A parabola or quadratic curve 
  3. An inverse graph 

How to Calculate the Slope or Gradient of a Graph

To find a slope of a graph, follow the following steps:

  1. On the graph, there is an x-axis and a y-axis
  2. They form a plane
  3.  The y-axis is on the vertical line of the graph
  4.  Also, the  x-axis is on the horizontal line of the graph
  5.  The y-axis is connected to the x-axis at the zero point
  6.   Also, the line that goes up, starting from zero is a positive y-axis
  7.   Additionally, the line that goes right, starting from zero is a positive X-axis
  8. We also have another line that moves down. This is a negative y-axis.
  9. From the left side of the plane, we have a negative X-axis
  10. Indicate your readings on the appropriate axis (y-axis and x-axis)
  11. Use a transparent ruler to connect the readings on the y-axis and the x-axis
  12. Apply the same ruler to draw a straight line between the two axis
  13. Ensure to apply pencil throughout your work
  14. Select two points on the line you have drawn with the ruler
  15. Connect the two lines to find your slope
  16. You need to know that Y = mx + c, where c is the intercept.
  17. This is how to calculate your slope by using [ Slope, S = (y2 – y1) / (x2 – x1)] 
  18. The plane is as (x , y)
  19. You can show coordinate points as M (x2 – x1) and N (y2 – y1)
  20. Subtract y1 from y2
  21. Also, subtract x1 from x2
  22. Divide (y2 – y1) by (x2 – x1) to get the slope 
  23. We also know slope as a gradient

How to Find an Intercept on a Graph

To find an intercept, you will have to extend the plotted line to cut through the y-axis or x-axis. The scale at which the plotted line cuts through any of the axis is the intercept of the graph. 

The graph should start from the origin at (0,0) for you to find the intercept. 

What are the Common Errors you Might Notice During Physics Practical

The common errors you need to avoid while carrying out physics practicals are:

  • Systematic Errors 
  • Random Errors 
  • Erratic Errors 
  • Zero Errors 
  1. Systematic Error is an error due to unforeseen circumstances like faulty lab apparatus and a blowing fan. 
  2. Random error is an error caused by poor judgment while reading the data.
  3. Erratic error is the type of error due to a lack of focus by the person doing the experiment. 
  4. Zero error is the type of error due to a bad pointer in an apparatus or faded calibration.

How to Solve Physics Practical Questions

To solve physics practical questions, follow these steps:

  1. Read Carefully: Understand the problem and what’s being asked. Note down given values.
  2. Identify Concepts: Figure out the relevant physics concepts involved, like motion, forces, electricity, etc.
  3. Formulas: Know the applicable formulas for the concepts. Match them with given values.
  4. Calculate: Plug values into the formulas and calculate the result.
  5. Units & Accuracy: Check units for consistency (e.g., meters, seconds). Round off to appropriate decimal places.
  6. Graphs (if needed): Plot data points on a graph. Choose suitable scales. Draw the best-fit line or curve.
  7. Interpretation: Analyze results. What do the numbers/graphs show? Draw conclusions.
  8. Error Analysis: Consider potential errors. Were measurements accurate? Calculate percentage error if needed.
  9. Compare & Discuss: If required, compare with theoretical expectations. Discuss any discrepancies.
  10. Conclusion: Summarize findings. State what you’ve learned or any improvements.
  11. Presentation: Neatly present the answer, including diagrams or graphs if applicable.
  12. Practice: Practice various problems to improve problem-solving skills.

Remember, physics practical questions often test your understanding of concepts, application of formulas, and analytical skills. Take your time, and step-by-step, you’ll become better at tackling these questions.

A sample of How to Conduct Physics Practical

Here is a sample of a question and answer on how to conduct physics practicals from West African Examination Council (WASSCE) – 2008

Question

You are provided with a uniform meter rule, a knife edge, masses, and other necessary apparatus.

  1. Suspend a meter rule horizontally on the knife edge. Read and record the point of balance G of the meter rule. keep the knife edge at this point throughout the experiment.
  2. Using the thread provided, suspend the object labeled W at the 15cm mark of the meter rule.
  3. Suspend a mass M of 20g on the other side of G. Adjust the position of the mass until the meter rule balances horizontally again.
  4. Read and record the position Y of the Mass M on the meter rule.
  5. Determine and record the distance L between the mass and G. Also, determine and record the distance D between W and G.
  6. Repeat the procedure for four other values of M = 30, 40, 50, and 60g. In each case, ensure that W is kept constant at the 15cm mark and the knife edge at G.
  7. Evaluate L-1 in each case. Tabulate your readings.
  8. Plot a graph of M on the vertical axis against L-1 on the horizontal axis
  9. Determine the slope, S, of the graph.
  10. Evaluate S/D
  11. State two precautions are taken to obtain accurate results.

Solution

Here is the solution to the above question

Aim of the practical

An experiment to determine the equilibrium of a meter rule with several masses.

Table for the experiment

where G = 49.50cm, D = 34.50cm

M(g)Y(cm)L(cm)L-1 (cm-1)
20.0099.2549.750.020
30.0084.0034.50.030
40.0075.3725.870.040
50.0070.2020.700.050
60.0066.7517.250.060
Table for meter rule experiment
how to conduct physics practical
Graph

Slope

To find the slope, we say

S = (y2 – y1) / (x2 – x1)

Therefore, the slope, S, is equal to

S = (60 – 20) / (0.05 – 0.02)

S = 1333.3gcm

To find S/D we now say

S/D = 1333/34.5 = 35

Precautions

  1. I ensure that all fans are switched off to avoid breeze from affecting the stability of the meter rule
  2. I ensured that the mass is properly hung on the required calibration.

What is Physics Practical Lab?

A physics practical lab is a hands-on learning space where students perform experiments to understand and apply concepts from physics. It’s like a playground for learning physics by doing.

Inside the lab, you’ll find tools and equipment like rulers, weights, and machines. Students follow instructions to set up experiments, measure things like distance and time, and record results.

They might test how things move, collide, or interact with forces. The lab helps students see physics in action and learn by trying things themselves. It’s a place where theories become real and exciting discoveries happen!

Frequently Asked Questions (FAQs) – Physics Practical

1. What is a physics practical?

A physics practical refers to hands-on experiments and activities conducted in a laboratory setting to explore and understand various concepts and principles of physics through direct observation and measurement.

2. Why are physics practicals important?

Physics practicals are crucial because they provide students with a tangible understanding of theoretical concepts. They enhance critical thinking, problem-solving skills, and allow students to apply theoretical knowledge in real-world scenarios.

3. What do physics practicals involve?

Physics practicals involve conducting experiments, making measurements, analyzing data, and drawing conclusions. These activities cover a wide range of topics, such as mechanics, optics, electricity, magnetism, thermodynamics, and more.

4. How are physics practicals conducted?

Physics practicals are typically conducted in a controlled laboratory environment. Students follow specific procedures outlined in the experiment instructions, use equipment and instruments, gather data, and often work in groups.

5. What equipment is used in physics practicals?

Physics practicals utilize a variety of equipment, including rulers, Vernier calipers, microscopes, pendulums, mirrors, lenses, resistors, capacitors, voltmeters, ammeters, and various specialized apparatus for specific experiments.

6. Are physics practicals safe?

Yes, safety is a top priority during physics practicals. Laboratories are equipped with safety measures, such as safety goggles, lab coats, and emergency equipment. Students are instructed on proper handling of equipment and materials to minimize risks.

7. Can physics practicals be done at home?

Some physics practicals can be adapted for home settings, but certain experiments may require specialized equipment or safety precautions that might not be feasible at home.

8. How do physics practicals relate to theory?

Physics practicals provide a concrete demonstration of theoretical concepts. They help students visualize and validate theoretical ideas, fostering a deeper understanding of the subject.

9. How are physics practicals evaluated?

Physics practicals are often evaluated based on the accuracy of measurements, the adherence to experimental procedures, the analysis of data, the quality of conclusions drawn, and sometimes through written reports or presentations.

10. Can physics practicals be challenging?

Yes, physics practicals can be challenging due to the precision required in measurements, the complexity of equipment, and the need to troubleshoot unexpected outcomes. However, they offer valuable learning experiences.

11. Are there any online resources for physics practicals?

Yes, there are various online platforms that provide virtual physics simulations and interactive experiments, allowing students to practice and learn physics concepts remotely.

12. Do physics practicals only apply to students pursuing physics degrees?

No, physics practicals are beneficial for students across various science and engineering disciplines as they promote skills like critical thinking, data analysis, and problem-solving that are applicable in many fields.

13. Can physics practicals inspire a career in physics?

Absolutely, engaging with physics practicals can spark an interest in pursuing a career in physics or related fields, as they provide a firsthand experience of the excitement and challenges of scientific exploration.

14. How can I excel in physics practicals?

To excel in physics practicals, focus on understanding the underlying concepts, practicing precise measurements, improving your data analysis skills, and seeking guidance from instructors or peers when needed.

15. Are there any famous physics experiments?

Yes, there are several iconic physics experiments like the double-slit experiment, the Michelson-Morley experiment, and the Millikan oil drop experiment, which have significantly contributed to shaping our understanding of the physical world.

Remember, these FAQs provide general information about physics practicals and their significance. Always refer to your specific course materials and instructor guidance for accurate details about your physics practical experiences.

Reference:

BBC

Capacitive Reactance: Definition, Formula, and Calculations

What is Capacitive Reactance?

Capacitive Reactance Definition: Capacitive reactance can simply be defined as the opposition to the flow of alternating current (a.c) in a circuit through a capacitor, and it is identified as Xc.

Capacitive reactance, often denoted as Xc​, is a phenomenon encountered when alternating current flows through a capacitor. A capacitor, composed of two conductive plates separated by an insulating material, stores electrical charge. When AC voltage is applied across a capacitor, it charges and discharges repeatedly, giving rise to capacitive reactance.

Capacitive Reactance
Capacitive Reactance

After connecting an alternating voltage of frequency (f), to a capacitor (C), this method will cause the alternating current (I) to flow in the opposite direction. Therefore, the voltage across the plates will lag on the current by 900.

Solved problem on capacitive reactance

Capacitive Reactance Formula

The capacitive reactance formula is Xc = 1 / (2πfC).

We can also use the formula, Xc = V0 / I0 as the capacitive reactance formula.

Where:

Xc = Capacitive reactance which is measured in ohms (Ω)

f = frequency and its si unit is in hertz (Hz)

C = Capacitance of the capacitor in farads (F)

I = Instantaneous current

I0 = Peak current

ω = angular speed (s.i unit = rads-1)

f = frequency

c = capacitance

θ = phase angle

It is important to know that angular frequency is equivalent to the angular speed.

Mathematical Formulation of Capacitive Reactance

Mathematically, capacitive reactance (Xc) is defined as the opposition offered by a capacitor to the flow of alternating current (AC).

It is inversely proportional to the product of the angular frequency (ω) of the AC signal and the capacitance (C) of the capacitor. Xc = 1 / (ω * C).

As the frequency of the AC signal increases, the capacitive reactance decreases, allowing more current to flow through the capacitor.

Capacitive Reactance Unit

Capacitive reactance, measured in ohms (Ω), is the resistance-like property that opposes the flow of alternating current (AC) through a capacitor in an electrical circuit.

Therefore, It increases as the frequency of the AC signal rises or the capacitance of the capacitor decreases. In simple terms, capacitive reactance is like a hurdle that AC electricity faces when trying to flow through a capacitor, getting higher as the AC signal gets faster or the capacitor gets smaller.

This concept helps us to understand how capacitors behave in AC circuits, affecting their ability to store and release electrical energy.

You may also like to read:

Dimension in Physics

How to Calculate Capacitive Reactance

To find capacitive reactance, you need to follow the following steps:

  • Read the Question
  • Extract your data from the question
  • Apply the formula to solve the problem
  • Don’t forget to add the unit (ohm) at the final answer

Here are problems on how to calculate and find capacitive reactance

Problem 1

An alternating current of 6mA and of frequency 60/π flow pass 3μF capacitor used in a radio circuit. What is the voltage across the capacitor?

Solution:

Data: Our first step is to read and understand the question so that we can easily extract our data.

Now the data is

Alternating current, I0 = 6mA = 6 x 10-3 A

frequency, f = 60/π

And the capacitance of the capacitor, c = 3μF = 3 x 10-6 F

we can clearly see that we now have Ic, f, and c

by using the formula for Xc, which is

Xc = 1 / 2Πfc

we can now substitute our data into the above equation to get

Xc = 1 / (2 x π x (60/π) x 3 x 10-6)

which will give us

Xc =

we will now have

Xc = 1 / 0.00036

Therefore, our reactive capacitance is equal to the

Xc = 2777.78 Ω

Now we can choose to write it in standard form as

Xc = 2.8 x 103 Ω

We can now apply the formula which says

Xc = V0 / I0

by making V0 subject of the formula, we have

V0 = I0 Xc

Thus,

V0 = 6 x 10-3 x 2.8 x 103

Therefore, the voltage across the capacitor can be written as

V0 = 16.8 V

We can also approximate it into

V0 = 17 V

Problem 2

At a frequency of 50Hz, a capacitor 5μF is connected to a circuit of 230V supply. Find the capacitive reactance.

Solution:

Data: from the above question, we can see that

Frequency, f = 50Hz

Capacitance, C = 5μF = 5 x 10-6F

Voltage, V = 230V

Now to calculate Xc, we apply the equation

Reactive capacitance, Xc = 1 / 2Πfc

Xc = 1 / ( 2x 3.142 x 50 x 5 x 10-6 )

Xc = 637 Ω

Problem 3

An Alternating current (a.c) supply of 240V is connected across a capacitor of 6μF and a resistance of 60 Ω in series at 70Hz. Find the capacitive reactance.

Solution

Data:

We have Capacitance, C = 6μF = 6 x 10-6F

Additionally, we can see from the question that the Frequency, f = 70Hz

And we know that the formula for reactive capacitance, Xc = 1 / 2Πfc

By substituting our data into the formula for reactive capacitance, implies that

Xc = 1 / ( 2x 3.142 x 70 x 6 x 10-6 )

we can now break it down into Xc = 1 / 0.00263928

Therefore, our final answer is, Xc = 379 Ω

Capacitive Reactance Dimensional Formula

The dimensional formula for capacitive reactance (Xc) is:

[Xc] = [Ω] = [F] / [Hz]

Where:

  • [Xc] represents the dimensional unit of capacitive reactance, which is ohms (Ω).
  • [F] stands for the dimensional unit of capacitance, which is farads (F).
  • [Hz] stands for the dimensional unit of frequency, which is hertz (Hz).

This formula shows that capacitive reactance is inversely proportional to both capacitance and frequency. In simpler terms, as the capacitance of a capacitor increases or the frequency of the AC signal decreases, the reactive capacitance (Xc) decreases, allowing more current to flow through the capacitor in an AC circuit.

Explanation: Understanding Reactive Capacitance

Here is a brief explanation of reactive capacitance.  As we have said earlier, reactive capacitance opposes the flow of current in an alternating-current circuit.

Therefore, reactive capacitance shares similarities with resistance due to the opposition of the flow of current. They are both measured in ohms (Ω).

We use Xc to indicate the symbol of reactive capacitance. While the letter X indicates reactance.

It is also important to know that the effect of resistance in the circuit is to affect the direct current (DC), while reactance helps in influencing the alternating current (a.c).

Reactance is also known as a property of an electric circuit element, this is because it opposes the flow of current in the circuit. 

Additionally, reactive capacitance is known as a property of capacitors. 

It’s also important to understand that it’s wrong to call capacitive reactance as capacitive resistance. 

When the frequency of a capacitance rises, it affects the capacitance. This process will make the reactive capacitance to be smaller. 

Understanding The Difference Between Capacitor and Resistance

Resistance 

Definition of Resistance: This is an opposition to the flow of charges(electrons) or current. The unit of resistance is in ohms. 

The unit of resistance which is in Ohm, originates from a German scientist known as George Simon Ohm. 

Ohm narrates the relationship between the current passing through a wire and the potential (p.d or V) between the ends of the wire.

He came up with a law that states:

“The current passing through a metallic conductor (for example a wire) is directly proportional to the potential difference provided the temperature remains constant”.

The equation for Ohm’s law is

I = VR 

Where I = current 

V = voltage 

And R = Resistance 

Resistance is not the same as resistivity or capacitive reactance. 

Capacitor 

Definition of Capacitor: A capacitor is a condenser that consists of two parallel metal conductors set apart by an insulator for storing electric charges. 

A capacitor is made up of two metal plates. The two plates are separated by a substance called an insulator (also known as dielectric), oil, or air. 

The si unit of capacitor is in Faraday. 

Capacitors permit the flow of more frequency current, they also resist the flow of low-frequency current in a circuit. 

Types of Capacitor

We have different types of capacitors and they are:

  • Fixed capacitor 
  • Variable capacitor 
  • Leyden Jar 
  • Electrolytic capacitor 

Note: Between the two parallel plates of a capacitor, there is a production of charges which is directly proportional to the potential difference. 

Capacitance of a capacitor: This is the quantity charge acquired when the potential difference between the two opposite plates is one volt. 

Basic Principles of Capacitive Reactance

Capacitors are fundamental elements in electronic circuits, capable of storing and releasing electrical energy. The charging and discharging processes within capacitors are the driving forces behind capacitive reactance.

As voltage is applied across a capacitor, it charges up, storing energy. Upon disconnection from the voltage source, the capacitor discharges, releasing the stored energy.

This cyclic behaviour is integral to capacitive reactance’s operation.

Effects of Capacitive Reactance

Capacitive reactance has several noteworthy effects on AC circuits:

  1. Phase Shift: In a capacitive circuit, current leads voltage by 90 degrees. This phase shift has practical applications in power factor correction.
  2. Current Limitation: As frequency rises, capacitive reactance diminishes, allowing more current to flow. This phenomenon can lead to overcurrent issues if not properly managed.
  3. Filtering Capacitors: Capacitors with high reactance at specific frequencies find use in filtering applications, allowing certain frequencies to pass while attenuating others.

What is the Effect of Frequency on Capacitive Reactance?

The effect of frequency on capacitive reactance is straightforward: as the frequency of the alternating current (AC) signal increases, the capacitive reactance decreases.

This means that a capacitor allows more current to flow through it at higher frequencies. Think of it like a gate that opens more easily for faster-moving traffic.

At lower frequencies, the capacitor’s opposition to the flow of current is stronger, acting like a narrower gate. This behaviour is crucial in AC circuits, as it influences the amount of current that can pass through a capacitor and how it behaves in different electronic applications.

Practical Applications

Capacitive reactance finds application in various fields:

  • Audio Systems: We use capacitors to separate bass frequencies in audio crossovers.
  • Touchscreens: The touch-sensitive surface of a touchscreen device utilizes the principles of capacitive reactance.
  • Power Factor Correction: Industries employ capacitors to offset reactive power, improving power factor and reducing energy costs.

Capacitive reactance finds application in a plethora of fields. In signal processing, capacitors with specific reactance values are employed in filters to allow or block certain frequency components.

Coupling and decoupling capacitors help isolate different stages of electronic circuits. Power factor correction, vital in efficient power distribution, utilizes capacitors to counteract reactive power and improve overall efficiency.

Furthermore, capacitive touchscreens and touch sensors rely on changes in capacitance to detect user interactions.

Characteristics and Properties of Capacitive Reactance

One fascinating characteristic of capacitive reactance is its frequency dependency. At lower frequencies, capacitors act as nearly open circuits, while at higher frequencies, they behave like short circuits.

Additionally, Xc introduces a phase shift of -90 degrees between the voltage and current waveforms.

This phase shift has implications for AC circuit analysis and design. When comparing AC and DC circuits, reactive capacitance (Xc) is a key factor that distinguishes their behaviours.

Limitations

While capacitive reactance offers numerous advantages, there are practical considerations to bear in mind. Capacitors have maximum voltage ratings that must not be exceeded to avoid breakdown.

Heating effects can occur due to the energy dissipated in the charging and discharging processes. Dielectric absorption can lead to temporary voltage retention after disconnection.

These factors must be carefully managed in circuit design and operation.

Comparison with Other Types of Reactance

Capacitive reactance stands alongside inductive and resistive reactance as key elements in circuit analysis. Inductive reactance arises from inductors, introducing a +90-degree phase shift.

Resistive reactance, which is proportional to resistance, remains constant regardless of frequency. In complex circuits, these reactance components combine to form the total impedance.

Experimental Methods and Measurement Techniques

To analyze reactive capacitance, engineers employ impedance analysis, frequency response measurements, and phase angle measurements.

These techniques provide insights into how capacitors behave in different circuit configurations and frequency ranges, facilitating accurate design and troubleshooting.

Troubleshooting and Problem Solving

In real-world applications, identifying and managing reactive capacitance-related issues is crucial. Understanding the effects of reactive capacitance on circuit behaviour helps engineers mitigate unwanted effects and optimize circuit performance.

Shielding, grounding, and careful component selection are practical approaches to address these challenges.

Future Trends and Developments

As technology advances, capacitive reactance continues to play a pivotal role. Advances in capacitor technology, such as materials and manufacturing techniques, can lead to improved performance and efficiency.

Integration with emerging technologies like Internet of Things (IoT) devices and wearable electronics holds promising potential for capacitive reactance application and exploration.

FAQs

What is the primary factor influencing capacitive reactance?

The frequency of the AC signal is the primary factor influencing reactive capacitance. As frequency increases, reactive capacitance decreases.

Can capacitive reactance cause circuit damage?

While reactive capacitance itself doesn’t cause damage, excessive current resulting from decreased reactance can harm components. Proper circuit design is crucial.

How does capacitive reactance compare to resistive reactance?

Reactive capacitance and resistive reactance both impede current flow but behave differently with respect to phase. Reactive capacitance leads voltage, while resistive reactance doesn’t introduce a phase shift.

Is capacitive reactance encountered only in AC circuits?

Yes, reactive capacitance is a phenomenon specific to AC circuits, as it involves the charging and discharging of a capacitor.

How do engineers manage capacitive reactance in circuits?

Engineers often use filtering capacitors and design circuits to control and utilize reactive capacitance effectively.

Are there situations where capacitive reactance is undesirable?

Excessive reactive capacitance can lead to unwanted phase shifts and overcurrent concerns. In some cases, it might need to be mitigated.

Reference

  1. wikipidea

Dimension in Physics

Dimension in physics
A table of dimension in of quantities in Physics

What is Dimension in Physics

The dimension of a physical quantity shows the application of the method of fundamental quantities like Mass (M), Length (L), and Time (T) to relate the fundamental units of a physical quantity. In dimension, we derive quantities by applying physical quantities. To understand dimension, we need to understand the method of measurements, fundamental quantities, derived quantities, and even their corresponding units in physics. This dimension helps us in finding the relationship between quantities. As we mentioned earlier, we apply Mass in the form of M, Length in the form of L, and Time in the form of T. Additionally, units like temperature will be in the form of K or θ, electric current in the form of (A) which is an expression for ampere, and finally luminous intensity as (II). Below are 2 videos explaining dimensions and their calculations.

Video 1

Video 2

Dimension in Physics

Table of Dimensions in Physics

QUANTITYDIMENSION
SpeedLT-1
VelocityLT-1
AccelerationLT-2
ForceMLT-2
Tensile stressML-1T-2
Elastic ForceMLT-2
Young ModulusML-1T-2
Work done in elastic spring/stringML2T-2
WorkML2T-2
Work done horizontalML2T-2
Work done verticalML2T-2
Kinetic energyML2T-2
Potential energyML2T-2
PowerML2T-3
Projectile time of flightT
Projectile maximum heightL
Projectile rangeL
Moment of forceML2T-2
DensityML-3
UpthrustMLT-2
Centripetal accelerationLT-2
Centripetal forceMLT-2
Linear velocityLT-1
Linear accelerationLT-2
PeriodT
FrequencyT-1
Period of simple pendulumT
MomentumMLT-1
Impulse
PressureML-1T-2
Work done by expanding gasML2T-2
WavelengthM
Wave velocityLT-1
Gravitational ForceMLT-2
Gravitational potentialL2T-2
Escape velocityLT-1
Electric forceMLT-2
Magnetic forceMLT-2
Dimension Table in Physics

You may also like to read:

SS1 Lesson Note: Introduction to Physics For First Term

How to Calculate Dimensions of Physical Quantity

To calculate the dimensions of a physical quantity, we need to be guided on the measurement of quantities in units of Length, Mass, and Time. These involve the dimensions Length (L), Mass (M), and Time (T) respectively. To find the dimension of any quantity, we substitute the units of the quantity into L, M, and T. You insert the symbols into a bracket in this way: [L], [M], and [T]. For example, to find the dimension of speed, we first write the formula as

Speed = Distance / Time

We know that the unit for Distance = meter (m) and meter is a length, which can be, Dimension of Distance = [L]

Note: We only have m which is [L]. If we have m2, we can now write [L2], and Time is in seconds which can also be T, Thus, the dimension of Time = T, To derive the dimension of speed, we say: Speed = Distance / Time. This implies that Speed = [L] / [T]. Therefore, the dimension of Speed = [LT-1] or [L][T-1].

Worked Examples on Dimension in Physics

Here are examples to make you understand dimension in physics: how to calculate dimensions of physical quantities

Example 1

  1. Find the dimension of an Area and Volume.

Solution:

For Area = Length x Breadth

Which shows that the dimension of Area = [L] x [L] = [L2]

Now, since volume = Length x Base x Height

The dimension of volume = [L] x [L] x [L] = [L3]

Example 2

2. Deduce the dimension of (a) force and (b) Pressure

Solution:

(a) force = mass x acceleration

which shows that

Dimension of force = dimension of mass x dimension of acceleration

and the Dimension of mass = [M]

while the Dimension of acceleration = Dimension of Velocity / Dimension of Time

and Dimension of Velocity = Dimension of Displacement [L] / Dimension of Time [T]

Thus, Dimension of Velocity = [L] / [T] = [L] [T-1] or [LT-1]

Therefore, by substituting the dimensions of velocity and time into acceleration,

we have

dimension of acceleration = [LT-1] / [T] = [LT-2]

Now, to finally derive the acceleration of force,

we say

Dimension of force = [M] x [LT-2] = [M] [LT-2] or [MLT-2]

(b) To find the dimension of pressure

We apply the formula pressure = force / area

which means Dimension of pressure = Dimension of force / Dimension of area

and Dimension of force = Dimension mass x Dimension of acceleration

you can remember that we earlier calculated the dimension of force as [MLT-2]

Also, Dimension of area = [L] x [L] = [L2]

Thus, Dimension of pressure = [MLT-2] / [L2]

This implies

Dimension of pressure = [ML-1T-2]

Example 3

Find the dimension of the following:

a. work

b. Velocity

c. Linear acceleration

d. Power

e. Pressure

f. Momentum

g. Impulse

Solution:

a. Work = Force x Distance

Dimension of Work = Dimension of Force x Dimension of Distance

Therefore, Dimension of Work = MLT-2 x L

This implies that, Dimension of work = ML2 T-2

b. Velocity = Displacement / Time = L / T = LT-1

c. Acceleration = Velocity / Time = LT-1 / T = LT-2

d. Power = work done / time = ML2 T-2 / T = ML2 T-3

e. Pressure = Force / Area = MLT-2 / L3 = ML-2T-2

f. Momentum = mass x velocity = M x LT-1 = MLT-1

g. Impulse = Force x Time = MLT-2 x T = MLT-1

Example 4

Find the dimension in the electrical system of the following:

a. Electric charge

b. Electrical Potential (voltage)

c. Capacitance (c)

Solution:

a. To find electric charge, we use the formula

electric charge = current (A) x time (T)

Thus, electric charge = [A] x [T] = [AT]

b. Electric potential (voltage) = power / current

Dimension voltage = ML2 T-3 / A = [MA-1L2 T-3]

c. Capacitance = charge / potential = [AT] / [MA-1L2 T-3]

Thus, capacitance = [M-1A2L-2 T4]

Reference

  1. Wikipedia

How to Find Resultant Force of Two Forces at an Angle

How to Find Resultant Force of Two Forces at an Angle

Forces are vectors that can influence the motion and equilibrium of objects. When two or more forces act on an object simultaneously, finding their resultant force becomes very important. Therefore, in this comprehensive guide, you will learn how to find the resultant force of two forces acting at an angle with a step-by-step guide and insightful tips.

a. Understanding Resultant Force

Before delving into the specifics, let us understand the concept of the resultant force. Resultant force refers to the single force that replaces multiple concurrent forces, producing the same effect on an object. It combines both magnitude and direction, capturing the net influence of the original forces.

b. Resolving Forces into Components

To calculate the resultant force, we must first resolve each force into its horizontal and vertical components. This process simplifies the vector addition, making it easier to find the resultant force.

c. Applying Trigonometry

With forces broken down into their components, we can now utilize trigonometric principles to calculate their magnitudes and angles accurately. This step plays a vital role in determining the resultant force.

d. Vector Addition – Finding the Resultant

Once the horizontal and vertical components are known, we can perform vector addition to find the resultant force. This involves adding the components along the respective axes and then combining them to yield the resultant force’s magnitude and direction.

Full Explantionation with Solved Problem

The formula below will help you understand how to find resultant force of two forces

R2 = F12+ F22

When two forces are acting at an angle, we will use the formula

R2 = F12+ F22 – 2 F1 F2 Cos ( 1800 – θ )

Where Forces F1 and F2 are acting at each other at an angle

How to Find Resultant Force of Two Forces at an Angle

Problem 1

Therefore, when you have two forces say 5N and 7N heading towards north ( the same ) direction and you are required to find the resultant of the two forces ( total force ). What you need to do is to add the two forces together to get the resultant force.

Forces F1 and F2 in the same direction where F1 = 5N and F2 = 7N

R = F1 + F2 = 5N + 7N = 12N

Forces F1, and F2 in the Same Direction While F3 is in Opposite Direction and F1 = 5N, F2 = 10N and F3 = 3N

R = F1 + F2 + F3 = 5N + (-3N) = 12N

F3 is in opposite direction and this is the reason we have (-3N) which shows that any force that is in negative direction is to be subtracted.

Problem 2

An object is acted upon by two forces of 5N and 12N. Calculate the resultant of the forces acting at an angle of 1200 to each other.

How do you find the resultant force of two forces at an angle?
How do you find the resultant force of two forces at an angle?

Solution

To calculate the resultant of the two forces, we need to extract our data from the above question

Data:

From the question above

We have two forces and they are as follows

F1 = 5N

F2 = 12N

The resultant ( R ) = ?

Angle between the two forces ( θ ) = 1200

Now, if we apply the formula

R2 = F12+ F22 – 2 F1 F2 Cos ( 1800 – θ )

Now, substitute the values of F1, F2 and θ into the above equation.

We now haveR2 = 52 + 122 – 2 x 5 x 12 Cos ( 1800 – 1200 )

which shows that R2 = 25 + 144 – 120 Cos 600

Hence, R2 = 169 – ( 120 x 0.5 )

We will now arrive at R2 = 169 – 60

After subtraction, we will have R2 = 109

By taking square root of each side we will have R = √109

And our final answer is R = 10.44N

Therefore, the resultant of the two forces acting between an angle of 1200 is Ten ( 10N ) newton.

Examples

a. Example Scenario: Forces at Right Angles

Let’s consider a scenario where two forces act at right angles to each other. This scenario simplifies the calculations, allowing us to demonstrate the process clearly.

b. Example Scenario: Forces at General Angles

In reality, forces often act at angles other than right angles. We will explore a scenario where two forces act at arbitrary angles and apply the same principles to find the resultant force.

Handling Equilibrium Situations

Sometimes, the resultant force may turn out to be zero, indicating that the object is in equilibrium. We will discuss the concept of equilibrium and explore scenarios where forces balance each other out.

Visualization Tools for Resultant Forces

To aid our understanding, we will introduce some visualization tools, such as vector diagrams and graphical representations. These tools can significantly simplify the process of finding the resultant force.

Practical Applications in Engineering

Understanding how to find the resultant force is crucial in various engineering applications. We will explore practical examples in structural engineering, mechanical systems, and more.

The Role of Angle Magnitude in Resultant Force

The angle between the forces significantly influences the magnitude of the resultant force. We will investigate how different angles affect the resultant force and what it means for the object in question.

Finding Resultant Force in Three Dimensions

In real-world scenarios, forces may act in three dimensions. We will expand our understanding to find the resultant force when forces lie in different planes.

Overcoming Challenges in Calculation

While the principles are relatively straightforward, certain scenarios can pose challenges in calculating the resultant force. We will identify common obstacles and explore techniques to overcome them.

Dynamic Forces and Time-Varying Resultants

In dynamic situations where forces change over time, the resultant force is no longer constant. We will address time-varying resultants and their significance.

Special Cases of Resultant Force

Some situations present unique cases of resultant forces. We will discuss special scenarios, such as concurrent forces passing through a single point, and how to handle them.

Analyzing Forces in Different Environments

Forces may behave differently in various environments, such as air, water, or space. We will examine how environmental factors can impact the calculation of resultant forces.

Tips for Accurate Measurement

Precision is crucial in calculating resultant forces, especially in critical applications. We will share some tips to ensure accurate measurements and reduce errors.

Verifying Results Experimentally

While calculations provide theoretical results, experimental verification is essential to validate our findings. We will discuss experimental techniques to verify the resultant force practically.

Real-World Examples and Case Studies

Delve into real-world examples and case studies where the knowledge of resultant forces played a pivotal role in solving practical problems.

Advancements in Resultant Force Analysis

With advancements in technology, computational tools have transformed how we analyze resultant forces. We will explore modern tools and software that aid engineers in their calculations.

How to Apply Resultant Force in Design

Learn how engineers and designers use knowledge of resultant forces to create safer and more efficient structures and systems.

Frequently Asked Questions

How to Find Resultant Force of Two Forces at an Angle?

To find the resultant force of two forces at an angle, follow these steps:

  1. Resolve each force into horizontal and vertical components using trigonometry.
  2. Add the horizontal components together and the vertical components together.
  3. Use the Pythagorean theorem to find the magnitude of the resultant force.
  4. Use inverse trigonometric functions to determine the angle of the resultant force.

What is the significance of finding the resultant force?

Finding the resultant force helps determine the net effect of multiple forces acting on an object, providing insights into its motion and equilibrium.

Can the resultant force be zero?

Yes, if the forces are balanced such that their combined effect cancels out, the resultant force can be zero, indicating equilibrium.

How does the angle between forces affect the resultant force?

The angle between forces influences the magnitude and direction of the resultant force. A larger angle can lead to a smaller resultant force, while a smaller angle can produce a larger resultant force.

How do engineers apply the concept of resultant force?

Engineers use the concept of resultant force to design structures, machines, and systems that can withstand and efficiently utilize forces in various applications.

Are there any software tools to calculate resultant forces?

Yes, there are specialized engineering software and computer-aided tools available that can efficiently calculate resultant forces in complex scenarios.

22. Conclusion

The knowledge of how to find the resultant force of two forces at an angle is a fundamental skill in physics and engineering. Therefore, by understanding the principles of vector addition, trigonometry, and resolving forces, you can confidently tackle any related problem. The ability to calculate resultant forces empowers you to create safer, more efficient, and innovative solutions.

You may also like to read

Refractive Index: Calculation of Apparent Displacement

Additionally, you can also check our websites

Wokminer – Search and apply for a job

Gas Laws in Physics

Concept of Gas Laws in Physics

Gas laws are a set of mathematical relationships that describe the behaviour of gases under various conditions. They provide a framework for understanding how gases respond to changes in temperature, pressure, volume, and the number of gas molecules. Gas is a state of matter that permeates our everyday lives. From the air we breathe to the fuel that powers our vehicles

gas laws
gas laws

2. The Basics of Gas Behavior

Gases consist of tiny particles, such as atoms or molecules, that are in constant motion and exhibit negligible intermolecular forces. These particles move randomly and collide with each other and the walls of their container, and thereafter exerting pressure on the walls of the container.

Video Explanation of Gas Laws Fomulas and Calculations

We can describe the behaviour of gases by using variables such as pressure (P), volume (V), temperature (T), and the number of moles (n) of gas molecules present. Gas laws provide relationships between these variables, which allow us to make predictions and understand how changes in one variable affect the other.

Gas Laws Formula

Here is a table of gas laws formulas along with their inventors, dates of invention, formulas, units, and respective dimensions based on the information available.

Gas LawInventorDate of InventionFormulaUnitDimensions
Boyle’s LawRobert Boyle1662PV = constantP * VPressure * Volume
Charles’s LawJacques Charles1787V / T = constantV / TVolume / Temperature
Gay-Lussac’s LawJoseph Gay-Lussac1802P / T = constantP / TPressure / Temperature
Avogadro’s LawAmedeo Avogadro1811V / n = constantV / nVolume / Amount of substance
Ideal Gas LawEmil Clapeyron1834 (further developed by others later)PV / nT = constantP * V / (n * T)Pressure * Volume / (Amount of substance * Temperature)
Combined Gas Law(P1 * V1) / (T1 * n1) = (P2 * V2) / (T2 * n2)VariesVaries
Gas Laws Formula Table

Properties of Gases

Gas is one of the states of matter. It is mostly obtained when a solid is heated to its melting point. Thereafter, its forces of attraction become weaker and will break into liquid. When we add more heat to the liquid to its boiling point, the liquid will start vaporizing and give up in the form of gases. Here are properties of gases you will need to know:

  1. The molecules possess a weak cohesive force
  2. The molecules are negligible.
  3. They fill the entire container in which they are.
  4. They have the highest kinetic energy compared to solid and liquid
  5. The degree of randomness is high.

From Kinetic Theory of Gases

According to the kinetic theory of gases, the following also apply:

  1. Gases are tiny particles that move in a straight line and collide with one another and also with the wall of the container.
  2. The collision between the gases is perfectly elastic.
  3. The cohesive force of the gases is also negligible
  4. The temperature of the gas is the measure of the average kinetic energy of the gas.

3. Boyle’s Law

Boyle’s Law, named after physicist Robert Boyle, states that the pressure of a gas is inversely proportional to its volume when the temperature and the number of gas molecules are constant. Mathematically, it can be expressed as: V ∝ (1/P)

Therefore, Boyle’s Law Formula is:

P1V1 = P2V2

where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume, respectively.

This law explains why squeezing a gas-filled balloon reduces its volume, causing an increase in pressure.

4. Charles’ Law

Charles’ Law, formulated by physicist Jacques Charles, describes the relationship between the temperature and volume of a gas when the pressure and the number of gas molecules are constant. According to Charles’s Law, as the temperature of a gas increases, its volume also increases, and vice versa.

Therefore, Charles’ Law states that the volume of a fixed gas is directly proportional to its absolute temperature, provided pressure is kept constant.

Mathematically, the charles’ law formula can be stated as: V ∝ T

Therefore, Charles’s Law Equation:

V1/T1 = V2/T2

Where V1 and T1 represent the initial volume and temperature, and V2 and T2 represent the final volume and temperature, respectively. As we can see, temperature and volume are involved in the case of Charles’ Law.

This law explains why a helium-filled balloon expands when exposed to heat.

5. Gay-Lussac’s Law

Gay-Lussac’s Law, named after chemist Joseph Louis Gay-Lussac, relates the pressure of a gas to its temperature when the volume and the number of gas molecules remain constant. According to this law, as the temperature of a gas increases, its pressure also increases, and vice versa. Mathematically, it can be expressed as:

Gay-Lussac’s Law Equation:

P1/T1 = P2/T2

where P1 and T1 represent the initial pressure and temperature, and P2 and T2 represent the final pressure and temperature, respectively.

This law helps in explaining why the pressure inside a car tire increases on a hot day.

6. The Combined Gas Law

The Combined Gas Law combines Boyle’s, Charles’s, and Gay-Lussac’s Laws into a single equation that describes the relationship between a gas pressure, volume, and a temperature. Thus, it allows us to calculate the changes in these properties when more than one variable is altered. The equation can be written as:

(P1V1)/T1 = (P2V2)/T2

The Combined Gas Law is particularly useful when studying gases undergoing complex changes.

7. Avogadro’s Law

Avogadro’s Law, proposed by the Italian scientist Amedeo Avogadro, states that equal volumes of gases, at the same temperature and pressure, contain an equal number of particles. This law helps us to understand the relationship between the volume and the amount (in moles) of gas. Mathematically, it can be expressed as:

Avogadro’s Law Equation:

V1/n1 = V2/n2

where V1 and n1 represent the initial volume and amount of gas, and V2 and n2 represent the final volume and amount of gas, respectively.

Avogadro’s Law is particularly relevant in stoichiometry calculations and understanding gas reactions.

8. Ideal Gas Law

The Ideal Gas Law combines all the previously discussed gas laws (Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s Laws) into a single equation that describes the behavior of an ideal gas. The Ideal Gas Law equation is given as:


PV = nRT

where (P) represents pressure, (V) represents volume, (n) represents the amount of gas (in moles), (R) represents the ideal gas constant, and (T) represents temperature in Kelvin.

The Ideal Gas Law allows us to relate the macroscopic properties of a gas to its microscopic behaviour.

9. Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas. Mathematically, it can be expressed as:

Dalton’s Law Equation:

Ptotal = P1 + P2 + P3 + ….

where Ptotal represents the total pressure of the gas mixture, and (P1 + P2 + P3 ) represents the partial pressures of each gas component.

Dalton’s Law is very important when it comes to dealing with gas mixtures, such as the composition of the Earth’s atmosphere.

10. Graham’s Law of Effusion and Diffusion

Graham’s Law of Effusion and Diffusion describes the rate at which gases effuse (escape through a small opening) or diffuse (mix with other gases).

According to Graham’s Law, the rate of effusion or diffusion of a gas is inversely proportional to the square root of Therefore, understanding these deviations is very important when studying gases in extreme conditions or when high precision is required.

12. Applications of Gas Laws in Everyday Life

Gas laws find practical applications in various fields and aspects of everyday life. Some notable examples include:

  • Understanding and predicting weather phenomena, such as changes in atmospheric pressure and temperature.
  • Designing and operating scuba diving equipment by considering gas solubility and pressure changes at different depths.
  • Developing efficient fuel combustion strategies in engines by optimizing air-fuel ratios.
  • Analyzing and interpreting gas data in industrial processes, such as chemical reactions and gas storage.

13. General Gas Law Formula

The general gas law in physics is the combination of Boyle’s, Charle’s, and Pressure laws for an ideal gas.

We can write the formula as

PV / T = Constant

Where

P = pressure

V = volume

T = Temperature

General Gas Law in Physics

And because of the constant, we can now write the formula for the general gas law as

P1V1 / T1 = P2V2 / T2

Similarly, the Ideal Gas Law is written as PV = nRT

GENERAL GAS LAW IN PHYSICS
GENERAL GAS LAW

14. General Gas Law in Physics: Solved Problems

Here are a few examples to help you understand how to apply the formula for general gas law and calculate a problem.

Problem 1: General Gas Law

A mass of gas at 170c and 700mmHg has a volume of 1.2m3. Determine its volume at 270c  and pressure at 750 mmHg.

Answer

Solution: To calculate the volume from general gas law, we need to first extract our data from the question

general gas law

Data:

Initial Pressure = P1 = 700mmHg

Final Pressure = P2 = 750mmHg

Initial Temperature = T1 = 170c = 273 + 17 = 290

Final Temperature = T2 = 270c = 273 + 27 = 300

Additionally, initial Volume = V1 = 1.2m3

Also, Final Volume = V2 = ?

Using the general gas law,

( P1 V1) / ( T1 )= ( P2 V2 ) / ( T2 )

By Making V2 subject of the formula,we have

Thus, V2 = ( P1 V1T2) / (P2 T1 )

by substituting the above expression with our data, we have:

Final volume, V2 = ( 700 x 1.2 x 300 ) / ( 750 x 290 )

Therefore, V2 = ( 252,000 ) / ( 217,500 )

This will give us our final answer as V2 = 1.2 m3

Therefore, the final volume is 1.2 m3

Problem 2: General Gas Law

The volume and pressure of a given mass of gas at 270C are 76 cm3 and 80 cm of mercury respectively. Calculate its volume at standard temperature and pressure (s.t.p)

Answer

general gas laws

Data

The original volume, V1 = 76 cm3

Initial pressure, P1 = 80 cmHg

Final pressure, P2 = 760 cmHg [Because of the standard temperature and pressure]

The initial temperature, T1 = 270C = 27 + 273 = 300 k

Final temperature, T2 = 273k [Because of the standard temperature and pressure]

Unknown

The final volume, V2 = ?

Formula

We will now apply our formula to solve the problem

( P1 V1) / ( T1 )= ( P2 V2 ) / ( T2 )

after inserting our data, we will now have

(80 x 76) / (300) = (76 x V2) / 273

By making V2 subject of the formula, we will obtain our answer as

V2 = 72.8 cm3

Problem 3: General Gas Law

A given mass of an ideal gas occupies a volume V at a temperature T and under a pressure P. If the pressure is increased to 2P and the temperature reduced to (1/2)T, then what is the percentage change in the volume of the gas?

Answer

The final answer to the above question is 75%

boyle's law

Explanation

Since we have a formula that says P1V1 / T1 = P2V2 / T2

and

V1 = V

P1 = P

P2 = 2P

T1 = T

T2 = (1/2)T

V2 = ?

We can make V2 subject of the formula and interpret it as

V2 = (P x V x (1/2)T ) / (2P x T)

And we will have

V2 = (1/2)V / 2 = 0.25 V

The percentage change in volume will now become

[(V1 – V2) / V1 ] x 100%

Which will now be

[(V – 0.25V) / V ] x 100%

hence

[0.75V / V ] x 100% = 0.75 x 100% = 75%

Therefore, the percentage change in the volume of the gas is 75%

Problem 4: General Gas Law

The pressure of a given mass of a gas changes from 300 Nm-2 to 120 Nm-2 while the temperature drops from 1270C to -730C. What is the ratio of the final volume to the initial volume?

Answer

The final answer to the question is V1 : V2 = 5 : 4

general gas laws

Explanation

P1 = 300 Nm-2

and P2 = 120 Nm-2

T1 = 127 + 273 = 400 K

we have T2 = -73 + 273 = 200 K

Formula

The formula we will use is P1V1 / T1 = P2V2 / T2

Solution

We will now insert our data into the formula

(300 x V1) / 400 = (120 x V2) / 200

To have a ratio of V1 to V2

Hence,

V2 : V1 = 60000 / 48000

This implies:

60 / 48 = V2 / V1

When we divide both 60 and 48 by 12, we will obtain

V2 / V1 = 5 / 4

Therefore, the ratio of V2 to V1 is 5 to 4

Problem 5: Boyle’s Law

The volume of a fixed mass of a gas is 200 cm3 at 270C. Calculate its temperature when the volume is reduced to (1/2) its original at the same pressure.

Answer

The final answer to the above question is -1230C

gas law - boyle's law calculation
boyle’s law calculation

Explanation

Data:

V1 = 200 cm3

T1 = 273 + 27 = 300K

V2 = (1/2)V1 = 0.5 x 200 = 100 cm3

Unknown:

T2 = ?

Formula:

V1 / T1 = V2 / T2

Solution

Applying our data into the formula, we will have:

200 / 300 = 100 / T2

By making T2 subject of the formula, we will have

T2 = (300 x 100) / 200 = 150 K

Hence, T2 = 273 K + t

t = 150 – 273 = -1230C

Therefore, the temperature is -1230C

Problem 6: Boyle’s Law

If the volume of a fixed mass of hydrogen in a container is 30 cm3 at a pressure of 50 mmHg. Find the volume of the gas if the pressure is 60 mmHg at constant pressure.

Answer

The final answer to the question is 25 cm3

Boyle's Law

Explanation:

We need to follow the following steps to solve the problem

First step (Data): The available information from the question

Initial volume, V1 = 30 cm3

Initial pressure, P1 = 50 mmHg

Final Pressure, P2 = 60 mmHg

Unknown: What we need to find

Final volume, V2 = ?

Formula: The equation that will help us to solve the problem

We will apply Boyle’s law, P1V1 = P2V2

Since we are to find the final volume, we will now make V2 subject of the formula

V2 = (P1V1) / P2

Therefore, we will use the above equation to solve the problem

Solution:

We will start by substituting our formula with available data from the question

V2 = (P1V1) / P2 = (50 x 30) / 60 = 25 cm3

Problem 7: Boyle’s Law

If the final volume of a gas is one-fifth of the initial volume. Find the ratio of the final pressure to the initial pressure at a constant temperature.

Answer

Data: Available information from the question

Final volume, V2 = (1/5)V1

Formula:

We will use Boyle’s law, P1V1 = P2V2

Solution

Changing our formula with the values in our data, we will obtain:

P1V1 = P2 x [(1/5)V1]

We can clearly see that V1 will cancel each other, leaving us with

5P1 = P2

Therefore, to find the ratio of the final to the initial pressure. We divide both sides by P1 to obtain:

P2 / P1 = 5

Thus, we can equally write the above question as

(P2 / P1) = (5 / 1)

Therefore, the ratio is

P2 : P1 = 5 : 1

Thus, the ratio of the final pressure to the initial is 5 to 1

Problem 8: Charles’ Law

The volume of a fixed mass of a gas is 200 cm3 at 270C. Calculate its temperature when the volume is reduced 1/2 its original at the same pressure.

Answer

The final answer to the question is -1230C

Explanation:

Data: The information available from the question

Initial volume, V1 = 200 cm3

Initial temperature, T1 = 270C = 27 + 273 = 300 K

Final volume, V2 = (1/2) V1 = 0.5 x 200 cm3 = 100 cm3

Unknown: What we need to find

Temperature = ?

Formula: The equation to solve the problem

V1 / T1 = V2 / T2

Solution:

Making T2 subject of the formula, and then substituting our data into the formula, we will have

T2 = (V2T1) / V1 = (100 x 300) / 200 = 150 K

Thus, since T2 = 273 + t

We will have:

150 = 273 + t

t = 150 – 273 = -1230C

Problem 9: Gay-Lussac’s Law

A fixed mass of air at standard temperature pressure is heated such that its volume remains unchanged. Calculate its tmeperature when its pressure is increased to 114 cm of mercury (Hg).

Answer

The final answer to the question is 136.50C

Explanation:

We will commence by following these important steps below:

Data: The information available from the question

Initial temperature, T1 = 00C = 273 K

The initial pressure, P1 = 76 cm Hg

Final Pressure, P2 = 114 cm Hg

Unknown: What we need to find

Final temperature, T2 = ?

Formula: The equation that will help us solve the problem

P1 / T1 = P2 / T2

Solution:

We will now make T2 subject of the formula and then insert our data into the formula

T2 = (P2T1) / P1 = (114 x 273) / 76 = 409.5

However, T = 273 + t

Therefore,

409.5 = 273 + t

Which implies that

t = 409.5 – 273 = 136.50C

Problem 10: Gay-Lussac’s Law

A mass of gas at a pressure of 50 mmHg is heated from 270C to 970C. If the volume is maintained at a constant. Calculate the pressure exterted by the gas.

Answer

The final answer to the above question is 61.67 mmHg

Explanation:

Data:

Initial pressure, P1 = 50 mmHg

The initial temperature, T1 = 270C = 27 + 273 = 300 K

Final temperature, T2 = 970C = 97 + 273 = 370 K

Unknown:

Final pressure, P2 = ?

Formula:

P1 / T1 = P2 / T2

Solution:

We will make P2 subject of the formula and then insert our data into the formula

P2 = P1T2 / T1 = (50 x 370) / 300 = 61.67 mmHg

Therefore, the pressure exterted by the gas is 61.67 mmHg

15. Measurement of Gas Pressure

U-tube manometer which contains water is used in measuring gas pressure. One end of the tube is connected to the gas supply, while the other end is opened to the atmospheric pressure. Therefore, when gas is supplied, pressure from the gas causes the water level to rise to a height h, on the other side of the tube. Thus, pressure at A and h are equal since are at the same horizontal level.

Pressure at A = Atmospheric pressure = pressure due to the volume of water at height h.

Also, the Pressure of gas = Atmospheric pressure + ρgh

Where

ρ = density

g = Acceleration due to gravity

h = height

Note: In 1643, an Italian scientist named Torricelli invented an instrument called a barometer for measuring the pressure of the air.

16. Summary

Summarily, we have seen that general gas laws form the backbone of our understanding of gas behaviour. Boyle’s Law, Charles’s Law, Gay-Lussac’s Law, Combined Gas Law, Avogadro’s Law, the Ideal Gas Law, Dalton’s Law, and Graham’s Law provide valuable insights into the properties and relationships of gases.

By applying these laws, we can make accurate predictions and solve real-world problems across various disciplines related to gas.

17. Frequently Asked Questions

Q1: What are the units of pressure in gas laws?
A1: Pressure in gas laws is typically measured in units such as Pascals (Pa), atmospheres (atm), or millimetres of mercury (mmHg).

Q2: How do gas laws relate to weather phenomena?
A2: Gas laws, such as changes in pressure and temperature, helps us in understanding and predicting weather phenomena like air masses, fronts, and atmospheric stability.

Q3: Can you provide an example of a real gas?
A3:Yes, gases such as carbon dioxide (CO2) and water vapor (H2O) display real gas behavior under certain conditions, deviating from the ideal gas model.

Q4: What happens to gas molecules at absolute zero?
A4: At absolute zero (-273.15°C or 0 Kelvin), gas molecules theoretically cease all motion and exhibit no pressure or volume.

You may also like to read:

How to Calculate Cubic Expansivity with Examples

Linear Expansivity: Definition and Calculations

Density of Water at 20 Degrees Celsius

Viscosity Definition

Surface Tension Definition

Also How to Calculate Specific Heat Capacity

And How to Calculate Relative Humidity Using a Formula

How to Calculate Cubic Expansivity with Examples

You can also check our website

Wokminer and search for any Job in any location

Reference

Tags

temperature and pressure, amount of gas, gas is directly proportional, gay lussacs, ideal gas law, remain constant, ideal gases, law of partial pressures, straight line, volume of the gas, kinetic theory, gas constant

Powered by WordPress & Theme by Anders Norén