Just the equations

Motion

Where:

v = speed

s = distance

t = time

Relevance:
These equations are used to calculate the speed of an object when the distance and time are known. It applies to motion scenarios like running, cycling, or vehicles traveling over a specific distance.

Example 1: Using the equation directly

A car travels 120 kilometers in 2 hours. Calculate its average speed in meters per second.
Convert 120 km to meters: 120 × 1000 = 120,000m
Convert time to seconds: 2 × 60 x 60 = 7200s

120,000 / 7200 = 16.7 m/s

Example 2: Rearranging to find time

A cyclist travels at an average speed of 5 m/s and covers a distance of 500 meters. Find how long it takes the cyclist.

Rearrange the equation: t = s/v

500 / 5 = 100 s

Where:

a = acceleration

v = velocity

t = time

Relevance:

This equation helps determine how quickly an object's velocity changes. Acceleration is seen in scenarios like cars speeding up, rockets launching, or objects in free fall.

Example 1: Using the equation directly

A car's velocity increases from 10 m/s to 30 m/s in 5 seconds. Find the acceleration.

Change in velocity (Δv) = 30 − 10 = 20 m/s

Change in time (Δt) = 5 - 0 = 5s

20 / 5 = 4 m/s².

Example 2: Rearranging to find time

A rocket accelerates at 10 m/s, and its velocity changes by 200 m/s. Find how long this takes.
Rearrange the equation: Δt = Δv / a

200 / 10 = 20 seconds

Mass and weight

Where:

W = weight

m = mass

g = gravitational field strength

Here are the requested details for the density, weight, and Hooke's Law equations:

Relevance: Weight is the force exerted by gravity on an object, and this equation is essential for calculating the force an object exerts due to gravity.

Example 1: Using the Equation Directly

An object has a mass of 10 kg. Calculate its weight on Earth (assuming g = 9.8 m/s).

10 × 9.8 = 98 N.

Example 2: Rearranging the Equation

An object weighs 150 N on Earth. What is its mass?

Rearrange the equation: m = W / g

150 / 9.8 = 15.31 kg.

Density

Where:

p = density

m = mass

V = volume

Relevance: Density measures how much mass is contained in a given volume of a substance, useful for identifying materials and predicting behavior in fluids.

Example 1: Using the Equation Directly

A cube of metal has a mass of 400 g and a volume of 100 cm³. Calculate the density of the metal.

400 / 100 = 4 g/cm³.

Example 2: Rearranging the Equation

A substance has a density of 3 g/cm³ and a volume of 50 cm³. What is the mass?

Rearrange the equation: m = ρ × V

3 × 50 = 150 g.

Effects of forces

Where:

F = force

k = spring constant

x = length

Relevance: Hooke's Law relates the force applied to a spring to the extension (or compression) of the spring, useful for understanding elastic materials.

Example 1: Using the Equation Directly

A spring has a spring constant of 200 N/m and is stretched by 0.5 m. Calculate the force exerted by the spring.

200 × 0.5 = 100 N.

Example 2: Rearranging the Equation

A force of 150 N is applied to stretch a spring. The spring constant is 300 N/m. What is the extension of the spring?
Rearrnage the equation: x = F / k

150 /300 = 0.5 m.

Where:

F = force

m = mass

a = acceleration

Relevance:
Newton's Second Law of Motion describes the relationship between force, mass, and acceleration. It applies to pushing objects, car crashes, and understanding forces acting on bodies.

Example 1: Using the equation directly

A box with a mass of 10 kg accelerates at 2 m/s². Calculate the force acting on it.

F = 10 × 2 = 20 Newtons.

Example 2: Rearranging to find mass

A force of 50 N causes an object to accelerate at 5 m/s. Find the object's mass.
Rearrange the equation: m = F / a

50 / 5 = 10 kilograms.

Turning effect of forces

Relevance: The moment of a force measures its turning effect about a pivot, crucial in mechanics and structural design.

Example 1: Using the Equation Directly

A 20 N force acts at a perpendicular distance of 0.5 m from a pivot. Calculate the moment of the force.

20 × 0.5 = 10 Nm.

Example 2: Rearranging the Equation

A force creates a moment of 15 Nm when applied at a perpendicular distance of 0.75 m from a pivot. Calculate the force.

Rearrange the equation: force = moment / distance

15 / 0.75 = 20 N.

Momentum

Where:

p = momentum

m = mass

v = velocity

Relevance: Momentum describes the motion of an object and its resistance to stopping, essential in collisions and dynamics.

Example 1: Using the Equation Directly

A car has a mass of 1,200 kg and travels at 20 m/s. Calculate its momentum.

1,200 × 20 = 24,000 kg·m/s.

Example 2: Rearranging the Equation

An object with momentum of 500 kg·m/s has a mass of 10 kg. What is its velocity?

Rearrange the equation: v = p / m

500 / 10 = 50 m/s.

Where:

F = force

t = time

m = mass

v = velocity

Relevance: Impulse measures the change in momentum due to a force acting over a time, critical in understanding collisions.

Example 1: Using the Equation Directly

A 10 N force acts on a ball for 2 seconds. Calculate the impulse.
Solution:

10 × 2 = 20 N·s.

Example 2: Rearranging the Equation

An impulse of 50 N·s is applied to an object over 5 seconds. What is the force?

Rearrange the equation: F = Δmv / Δt 

50 / 5 = 10 N.

Where:

F = force

p = momentum

t = time

Relevance: This equation relates force to the rate of change of momentum, used in collision analysis and dynamics.

Example 1: Using the Equation Directly

A car's momentum changes by 2,000 kg·m/s over 5 seconds. Calculate the force.

2,000 / 5 = 400 N.

Example 2: Rearranging the Equation

A force of 250 N acts on an object for 4 seconds. What is the change in momentum

Rearrange the equation: Δp = F x Δt

250 × 4 = 1,000 kg·m/s.

Energy

Where:

Ek = kinetic energy

m = mass

v = velocity

Relevance: Kinetic energy measures the energy of an object in motion, key in analyzing energy transformations.

Example 1: Using the Equation Directly

A 2 kg object moves at a velocity of 3 m/s. Calculate its kinetic energy.

1/2 × 2 × 3² = 9 J.

Example 2: Rearranging the Equation

An object has 25 J of kinetic energy and a mass of 4 kg. What is its velocity?

Rearrange the equation: v = sqrt(2Kem)

sqrt(2 × 25 x 4) = 3.54 m/s.

Where:

Ep = gravitational potential energy

m = mass

g = gravitational field strength

h = height

Relevance: This equation calculates the energy stored in an object due to its height in a gravitational field, crucial in energy conservation problems.

Example 1: Using the Equation Directly

A 5 kg object is lifted to a height of 2 m. Calculate its gravitational potential energy (g = 9.8 m/s²).

5 × 9.8 ×2 = 98 J.

Example 2: Rearranging the Equation

An object has 196 J of gravitational potential energy and a mass of 10 kg. What is its height?

Rearrange the equation: h = Ep / mg 

196 / 10 × 9.8 = 2 m.

Work

Where:

W = work done

F = force

d = length

E = energy

Relevance:
Work done is the energy transferred when a force moves an object over a distance. It applies to lifting objects, moving vehicles, and machines doing work.

Example 1: Using the equation directly

A person pushes a box with a force of 100 N over a distance of 5 meters. How much work is done?

100 × 5 = 500 Joules.

Example 2: Rearranging to find force

A crane does 1200 J of work to lift an object 6 m. Find the force applied.
Rearrange the equation: F = W / F

1200 / 6 = 200 Newtons.

Energy resources

Relevance:
Efficiency measures how effectively energy is converted into useful work. This applies to machines, engines, and electrical appliances.

Example 1: Using the equation directly

A machine has an energy input of 500 J and a useful output of 300 J. Find its efficiency.

300 / 500 × 100 = 60% efficient.

Example 2: Rearranging to find useful energy output

If a machine's efficiency is 80% and the energy input is 1000 J, find the useful energy output.
Rearrange the equation: Efficiency × Input / 100

80 × 1000 / 100 = 800 Joules.

Power

Where:

P = power

W = work done

t = time

Where:

P = power

E = energy

t = time

Pressure

Where:

p = pressure

F = force

A = area

Where:

p = pressure

p = density of liquid

g = gravitational field strength

h = height/depth

Thermal physics

Kinetic particle model of matter

Thermal properties and temperature

Waves

General properties of waves

Light

Electricity and magnetism

Electrical quantities

Electrical circuits

Electromagnetic effects

Space physics

The earth

The universe