The following page is a list of formulas and very brief explanations.

**Density**is the ratio of a substance's mass to its volume and can be expressed mathematically as

whereD=M/V

*D*is density,

*M*is mass, and

*V*is volume.

Example:

What is the density of an object having a mass of 8 kg and a volume of 2 cubic meters?

D = M/V

D = 8/2

D = 4 kg/m^{3}

**Temperature**:

In science, we will use Celsius or Kelvin temperature scales to describe temperature.

To convert between Celsius and Kelvin:

Kelvin = Celsius + 273.15

Celsius = Kelvin - 273.15

To convert between Celsius and Fahrenheit:

Fahrenheit = Celsius * (9/5) + 32

Celsius = (Fahrenheit - 32) * 5/9

**Gas Laws:**

*Charles's Law*

The volume of a gas is directly proportional to its temperature in kelvins if the pressure and number of molecules are constant.

V1T1=V2T2

*Boyle's Law*

The volume of a gas is inversely proportional to its pressure if the temperature and the number of molecules are constant.

P1V1=P2T2

*Combined Gas Law*

Pressure is inversely proportional to volume, or higher volume equals lower pressure. Pressure is directly proportional to temperature, or higher temperature equals higher pressure.

Example:(P1V1)/T1=(P2V2)/T2

If a sample of gas initially has a pressure of 2 atm, a volume of 3 liters, and a temperature of 300 K, what would its final volume be if the pressure changed to 1.5 atm and the temperature changed to 290 K?

(P1V1)/T1=(P2V2)/T2 (2 • 3)/300 = (1.5 • V)/290

6/300 = 1.5 V/290

290 • (6/300) = 1.5 V

5.8 = 1.5 V

5.8/1.5 = V

3.87 l = V

*Ideal Gas Law*

When the number of molecules are included in calculations, the following formula can be used:

PV = nRT

This law can be converted into the following form which will allow memorization of only 1 formula for gas law problems:

P1V1 P2V2

_____ = _____

n1T1 n2T2

**Motion**:

Finding final velocity:

wherevf = vi + at

*vf*is final velocity,

*vi*is initial velocity,

*a*is acceleration, and

*t*is elapsed time.

Example:

An object is moving at a rate of 3 m/s and accelerates at a rate of 2 (m/s)/s for 5 seconds. What is its final velocity?

vf = vi + at

vf = 3 + 2 • 5

vf = 3 + 10

vf = 13 m/s

Finding average velocity:

wherev(ave) = (vf+vi)/2

*v(ave)*is average velocity,

*vf*is final velocity, and

*vi*is initial velocity. Also:

v(ave) = ((vi + at) + vi)/2

Finding final position (final distance from a reference point):

df = di + vit + 1/2at^{2}

wheredfis the final, total displacement…

diis the initial displacement. (How far from whatever point of reference is the object when the thing starts accelerating?)…

viis the initial velocity of the object at the beginning of the acceleration.

tis the elapsed time from the beginning of the acceleration until the end of the period being observed.

(vitaccounts for the motion of the object based on its starting velocity. It keeps covering distance at the initial rate, and additionally, it accelerates and covers more distance.)

ais the acceleration andtis elapsed time.

Example:

An object begins 10 meters from a mark on a track with an initial velocity of 3 m/s. If it accelerates at a rate of 5 (m/s)/s for 4 seconds, how far from the mark does it end up?

df = di + vit + 1/2at^{2}

df = 10 + 3(4) + 1/2(5)(4)^2

df = 10 + 12 + 1/2 (5) (16)

df = 10 + 12 + 40

df = 62 m

**Force**:

Where

*F*is force,

*a*is acceleration, and

*m*is mass, then:

F = ma

**Force of Friction**—friction always opposes forces that are moving (or trying to move) an object.

Ff = Fnμ

whereFfis the force of friction,Fnis the normal force, andμis the coefficient of friction (a number that is looked up).

The normal force is the portion of weight that is perpendicular to the surface. For a flat surface (the angle between the surface and the horizon is zero, θ = 0):

Fn = mg

where m is mass and g is acceleration due to gravity (9.8 (m/s/)s on earth)

**Work and Energy**:

**Work**is found by

whereW = Fd

*W*is work,

*F*is force applied (not net force!), and

*d*is displacement/distance.

**Kinetic energy**(KE) is found with this equation:

whereKE = 1/2 mv^{2}

*KE*is kinetic energy,

*m*is mass and

*v*is velocity.

The

**potential gravitational energy**can be found with this equation:

wherePE = mgh

*PE*is potential gravitational energy,

*m*is mass,

*g*is acceleration due to gravity, and h is height.

On earth, acceleration due to gravity is 9.8 (m/s)/s

The amount of

**elastic potential energy**is determined by how hard it is to compress or stretch something and how far it is stretched or compressed.

The equation to find this is:

wherePE = 1/2kd^{2}

*PE*is potential elastic energy,

*k*is a constant specific to a particular stretchy thing (spring, rubber band, etc.) and d is the distance that it is stretched or compressed (sometimes x is used instead of

*d*, as in the illustration)

**Heat/Energy Transfer**

Now to find heat, we can use the formula:

WhereQ = mcΔT

*Q*is heat or thermal energy,

*m*is mass,

*c*is a number called specific heat that you either look up or calculate, and

*ΔT*is the change in temperature.

EXAMPLES:

How much heat is absorbed by 200 grams of water that starts at 25C and ends up at 30C, given that the specific heat of water is 4.2 J/g°C?Q = mcΔTQ = 200•4.2•(30-25)Q = 200•4.2•5Q = 4200 JWhat is the specific heat of a metal that has a mass of 50 grams and changes temperature from 100C to 30C and gives off 4200 J of thermal energy?Q = mcΔT4200 = 50•c•(100-30)4200 = 50•70•c4200 = 3500c4200/3500 = c1.2 = c

Given the following information, find the work done on a 6.5 kg object after 4 seconds:

A = 16 N

B = 4 N

C = 14 N

D = 9 N

B = 4 N

C = 14 N

D = 9 N

STEP 1: Find Net Force by resolving the UpDown forces, resolving the LeftRight forces, and then using the Pythagorean Theorem:

F(net)

^{2}= UpDown

^{2}+ LeftRight

^{2}

^{}F(net)

^{2}= (16 - 4)

^{2}+(14 - 9)

^{2}

F(net)

^{2}= (12)

^{2}+ (5)

^{2}

F(net)

^{2}= 144 + 25

F(net)

^{2}= 169

**F(net) = 13 N**

STEP 2: Find Acceleration where the force is the net force on the object:

F = ma

13 = 6.5a

13/6.5 = a

**2 (m/s)/s = a**

STEP 3: Find the distance through which the force acted due to the net force.

df = 1/2at

^{2}

df = 1/2 • 2 • 4

^{2}

**df = 16 m**

STEP 4: Find the work done by the net force through the calculated distance.

W = Fd

W = 13 • 16

**W = 208 J**

**To find the**

__final velocity__in the above:Do Step 1 above.

Do Step 2 above.

STEP 3:

Vf = at

Vf = 2 • 4

**Vf = 8 m/s**

**To find**

__final kinetic energy__, first find the final velocity (above), and then:KE = 1/2mv

^{2}

KE = 1/2•6.5•8

^{2}

KE = 1/2 • 6.5 • 64

**KE = 208 J**