science:physics

# physics

## Introduction

## a few basic equations

**velocity**- velocity = distance travelled / unit time
- 100kph = 100kph x 1000m / (3600secs) = 27.8m/s

**acceleration**- acceleration = velocity change / unit time
- a car accelerating uniformly from 0 to 100kph in 10secs will have:
- average acceleration = 100kph x 1000m / (3600secs x 10secs) = 2.78m/second
^{2}

**momentum**- linear momentum = mass x velocity
- angular momentum = the moment of inertia (I) x angular velocity (ω)
- moment of inertia (I) = mass x radius
^{2}where radius is distance to the centre of motion- hence a ballerina will spin faster when limbs are close to the body as moment of inertia is reduced and thus angular velocity must increase as momentum remains constant

- NB. momentum is conserved during collisions, explosions, and other events involving objects in motion, however, in the real world, some kinetic energy is lost to heat during a collision, especially if the collision is inelastic, and thus overall momentum will fall.
- net system momentum is constant if the net external force (or for angular momentum, external torque) is zero

**force**- force = mass x acceleration = change in momentum / unit time
- 1 newton (N) = 1 kilogram X meter/second
^{2} - lifting a 1kg object to overcome gravity requires a force of 1kg x acceleration due to gravity = 9.8N
- force to move a 2000kg car from 0 to 100kph in 10secs = 2000 x 2.78 = 5556N

**kinetic energy**- energy = 1/2 x mass x velocity
^{2} - 1 Joule = 1 kg m
^{2}/sec^{2}

**gravitational potential energy**- energy = mass x gravitational constant x height

**work**- work = force x distance = change in kinetic energy
- 1 Joule = 1Nm
- work done to lift a 10kg object 1m off the ground = 10kg x 9.8 x 1m = 100J
- work done to accelerate a 2000kg car from 0 to 100kph (excl. frictional and wind resistance work) = 0.5 x 2000 x (27.8m/s)
^{2}= 27,800J

**power**- power = work / time taken = energy used / time
- 1W = 1 Joule/sec
- power to accelerate a 2000kg car from 0 to 100kph in 10sec = 27,800J / 10sec = 2780W
- note that combustion engines are only about 40% efficient so the engine would actually need to use 2780/0.4 =7kW of power
- petrol has a energy density of 44MJ/kg thus over 10sec at 7000W requires 70000J = 70kJ and thus ~44000/70 = 1.6mL of fuel IF there were no frictional forces to overcome such as wind resistance, etc.

**energy capacity of a battery**- capacity is measured in Watt-hours (Wh) ie. how many watts of power is available over 1 hour

science/physics.txt · Last modified: 2022/07/31 00:12 by gary1