Math Skill #1 Conversions
Conversions can be made by multiplying any given amount by a correct
conversion factor. A conversion factor is any fraction in which 2 different
units and amounts are equal to each other.
Examples of Conversion Factors: 1 day/24 hours 1000 m / 1 km
12 cans / 1 dozen 100 years / 1 century
The metric system is used in science because it is based on the number 10.
You can move the decimal left or right to the correct unit or multiple by a
correct conversion factor.
Metric System Basic Units / Amounts
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kilo hecto Deka Base deci centi milli
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k h Da (1) d c m
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1,000 100 10 .1 .01 .001
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g (grams) mass
m (meters) distance
L (liters) volume
s (seconds) time
Examples of Metric Conversions:
1000m
To convert .345 km to m .345 km X ______ = 300 m
1 km
** note that units showing in the numerator and denominator will cancel
each other out. This will leave the desired unit for the answer.
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Math Skill #13 Velocity
SPEED EQUATION: SPEED = DISTANCE / TIME or v = d/t
Example 1: What is the velocity in meters per second of a swimmer who
swims exactly 110 m toward the shore in 72 s?
v = d/t
v = 110 m / 72 s
v = 1.5 m/s toward the shore
** velocity includes speed in a direction
Example 2: Calculate the distance in meters a cyclist would travel in
5.0 hours at an average velocity of 12.0 km/h to the
southwest?
d = vt
d = (12.0 m/s)(5.00 h)
d = (60.0 km)(1000 m)= 60,000 m
** we had to multiple the 60 km by 1,000 b/c the question asked
for meters.
Example 3: Calculate the time in seconds an olymic skier would take to
finish a 2.6 km race at an average velocity of 28 m/s downhill?
t = d/v
t = (2.6 km)(100 m/km) ** must convert to meters
t = 26,000 m / 28 m/s
t = 93 s
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Math Skill #14 Momentum
MOMENTUM EQUATION: MOMENTUM = (MASS)(VELOCITY) or p = mv
Example 1: Calculate the speed of a 75 kg skater moving forward at 16 m/s.
p = mv
p = (75 kg)(16 m/s)
p = 1200 kg x m/s forward
Example 2: What is the velocity of a 1.35 kg baseball with a momentum of
5:06 kg x m/s?
v = p/m
v = 5.06 kg x m/s / 1.35 kg
v = 3.75 m/s
Example 3: Calculate the mass of a man jogging down the road at 2.65 m/s
with a momentum of 230 kg x m/s.
m = p/v
m = 230 kg x m/s / 2.65 m/s
m = 85 kg
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Math Skill #15 Acceleration
ACCELERATION EQUATION: FINAL VELOCITY - INITIAL VELOCITY
ACCELERATION = ------------------------------
TIME
Example 1: A turtle swimming in a straight line toward shore has a speed
of 0.50 m/s. After 4.0 s, its speed is 0.80 m/s. What is the
turtle's average acceleration?
a = Vf - Vi / t
a = (.80 m/s - .50 m/s) / 4.0 s
a = 0.075 m/s squared toward the shore
Example 2: Marisa's car accelerates at an average rate of 2.6 m/s squared.
Calculate how long it takes her car to accelerate from 24.6 m/s
26.8 m/s.
t = v / a
t = (26.8 m/s - 24.6 m/s) / 2.6 m/s squared
t = 0.85 s
Example 3: A cyclist travels at a constant velocity of 4.5 m/s, then speeds
up with a steady acceleration of 2.3 m/s squared. Calculate the
cyclist's speed after accelerating for 5.0 s.
Vf = Vi + at
Vf = 4.5 m/s + (2.3 m/s squared)(5.0 s)
Vf = 16 m/s
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Math Skills #16 Newton's 2nd Law of Motion or Force
FORCE EQUATION: FORCE = (MASS)(ACCELERATION) OR F = ma
Example 1: What is the net force necessary for a 1,600 kg automobile to
accelerate forward at 2.0 m/s squared?
F = ma
F = (1,600 kg)(2.0 m/s squared)
F = 3,200 N
** the unit for force is the Newton or N.
Example 2: A baseball accelerates downward at 9.8 m/s squared. If the
gravitational force acting on the baseball is 1.4 N, what is the
baseball's mass?
m = F / a
m = 1.4 N / 9.8 m/s squared
m = 0.14 kg
Example 3: A sailboat and its crew have a combined mass of 655 kg. If the
sailboat experiences an unbalanced force of 895 N pushing it
forward, what is the sailboat's acceleration?
a = F / m
a = 895 N / 655 kg
a = 1.37 m/s squared in the direction of the force.
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Weight equals mass times free-fall acceleration.
**The force on an object due to gravity is called its weight.
WEIGHT EQUATION: WEIGHT = MASS X FREE FALL ACCELERATION OR w = mg
Example 1: What is the weight of Coach Hawkins if he has a mass of 110 kg?
w = mg
w = (110 kg)(9.8 m/s squared)
w = 1,078 N
** note that g is always 9.8 m/s squared
Example 2: Calculate the mass of a meteor that has a weight of 2,600 N.
m = w / g
m = 2,600 N / 9.8 m/s squared
m = 265.3 kg
*** note that we do not solve for acceleration due to gravity because it is
a constant value of 9.8 m/s squared.
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Math Skills #17 Work
Work is defined as a quantity that measures the effects of a force acting
over a distance.
WORK EQUATION: work = force x distance or W = F x d
Example 1: A crane uses an average force of 5200 N to lift a girder 25 m.
How much work does the crane no on the girder?
W = F x d
W = (5200 N)(25 m)
W = 130,000 J
** note that the unit for work is the joule or J.
Example 2: You must exert a force of 4.5 N on a book to slide it across a
table. If you do 2.7 J of work in the process, how far have you
moved the book?
d = W / F
d = 2.7 J / 4.5 N
d = 0.6 m
Example 3: A child pulls a sled up a snow-covered hill. In the process, the
child does 405 J of work on the sled. If she walks a distance of
15 m up the hill, how large a force does she exert on the sled?
F = W / d
F = 405 J / 15 m
F = 27 N
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Math Skills #18 Power
Power is defined as a quantity that measures the rate at which work is done.
POWER EQUATION: Power = work / time or P = W / t
Example 1: While rowing in a race, John does 3960 J of work on the cars in
60.0 s. What is his power in watts?
P = W / t
P = 3960 J / 60.0 s
P = 66.0 W
Example 2: The world's most powerful tugboats, which are built in Finland
are capable of providing 8,170,000 W of power. How much work
does one of these tugboats do in 12.0 s?
W = Pt
W = (8,170,000 W)(12.0 s)
W = 98,000,000 J
Example 3: A runner exerts a force of 334 N against the ground while running
a distance of 50.0 m. The runner's power output over this
distance is 3.71 kW. How much time does it take the runner to
travel 50.0 m?
t = W / P
t = (50.0 m)(334 N) / 3,710 W
t = 4.50 s
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Math Skills #19 Mechanical Advantage
Mechanical Advantage is a quantity that measures how much a machine multiplies
force or distance.
MECHANICAL ADVANTAGE EQUATION: MA = output force / input force
MA = input distance / output distance
Example 1: Calculate the mechanical advantage of a ramp that is 6.0 m long
and 1.5 m high.
MA = input distance / output distance
MA = 6.0 m / 1.5 m
MA = 4.0
Example 2: Alex pulls on the handle of a claw hammer with a force of 15 N.
If the hammer has a mechanical advantage of 5.2, how much force
is exerted on a nail in the claw?
output force = (MA)(input force)
output force = (5.2)(15 N)
output force = 78 N
Example 3: While rowing in a race, Bob pulls the handle of an oar 0.80 m on
each stroke. If the oar has a mechanical advantage of 1.5 how far
does the blade of the oar move through the water on each stroke?
output distance = input distance / MA
output distance = 0.80 m / 1.5
output distance = 0.53 m
** note that mechanical advantage does not have a unit. It is the number of
times a machine multiplies the force put into it.
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Math Skills #20 Gravitational Potential Energy
GPE is defined as the stored energy resulting from the relative positions of
objects in a system.
GRAVITATIONAL POTENTIAL ENERGY EQUATION: GPE = mass x gravity x height
GPE measured in (J)joules gravity represented by 9.8 m/s squared
mass measured in (g)grams height measured in (m)meters
** g, is always 9.8 m/s squared. It may not be mentioned in some word
problems. You must understand to put it in the formula as needed.
Example 1: What is the GPE of a car with a mass of 1200 Kg at the top of a
42 m high hill?
GPE = mgh
GPE = (1200 kg)(9.8 m/s squared)(42 m)
GPE = 490,000 J
Example 2: A student holds a .055 kg egg out a window. Just before the
student releases the egg, the egg has 8.0 J of GPE with respect
to the ground. How high is the student's arm from the ground?
h = PE / mg
h = 8.0 J / (0.055 kg)(9.8 m/s squared)
h = 15 m
Example 3: A driver has 3400 J of GPE after steppin up onto a driving
platform that is 6.0 m above the water. What is the driver's
mass in kilograms?
m = PE / gh
m = 3400 J / (9.8 m/s squared)(6.0 m)
m = 58 Kg
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Math Skills #21 Kinetic Energy
Kinetic energy is the energy of a moving object due to its motion.
KINETIC ENERGY EQUATION: 1/2 x mass x speed squared
Example 1: Calculate the kinetic energy in joules of a 1500 kg car moving
at 18 m/s.
KE = (1/2)(m)(velocity)squared
KE = (1/2)(1500 kg)(18 m/s)squared
KE = 243,000 J
Example 2: A 35 kg child has 190 J of kinetic energy after sledding down
a hill. What is the child's speed in m/s at the bottom of the
hill?
v = the square root of 2(Kinetic energy) / mass
v = the square root of 2(190 J) / 35 kg
v = 3.3 m/s
Example 3: A bowling ball traveling 2.0 m/s has 16 J of kinetic energy. What
is the mass of the bowling ball in kilograms?
m = 2(Kinetic Energy) / velocity squared
m = 2(16 J) / 2 m/s squared
m = 8.0 kg
** pay close attention to order of operations when working KE problems.
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Math Skills #22 Efficiency
Efficiency is a quantity, usually expressed as a percentage, that measures
the ratio of useful work output to work input.
EFFICIENCY EQUATION: efficiency = useful work output / work input
Example 1: Two boys calcualte that they must do 1800 J of work to push a
piano up a ramp. However, because they must also overcome
friction, they must actually do 2400 J of work. What is the
efficiency of the ramp?
efficiency = work output / work input
efficiency = (1800 J)(2400 J)
efficiency = 0.75 or 75%
Example 2: It takes 1200 J of work to lift the car high enough to change a
tire. How much work must be done by the person operating the jack
if the jack is 25% efficient?
work input = useful work output / efficiency
W in = (1200 J)(0.25)
W in = 4800 J
Example 3: A windmill has an efficiency of 37.5%. If a gust of wind does
125 J of work on the blades of the windmill, how much output work
can the windmill do as a result of the gust?
Work Output = (efficiency)(work input)
W out = (0.375)(125 J)
W out = 46.9 J
*** remember that when you work with a percentage in the problems make sure
that you us it in decimal form. Example .25 should be used in the math
equations to represent 25%.
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Math Skills #23 Temperature Conversions
Temperature is a measure of the average kinetic energy of all the particles
within an object.
Converting Celcius to Fahrenheit: Tf = 9/5 (Tc + 32)
Example 1: Convert the boiling point of hydrogen (-252.87 C) to Fahrenheit.
Tf = 9/5 (Tc + 32)
Tf = 9/5 (-252.87 + 32)
Tf = -423.2 F
Converting Fahrenheit to Celcius: Tc = 5/9 (Tf - 32)
Example 2: The temperature of air in the desert is 110 F. Convert this
amount to degrees celcius.
Tc = 5/9 (Tf - 32)
Tc = 5/9 (110 - 32)
Tc = 43 C
Converting Celcius to Kelvin: T = Tc + 273
Example 3: The air in a typical living room is 21 C. Convert to Kelvin.
T = Tc + 273
T = 21 C + 273
T = 294 K
** Remember that you can convert celcius to Kelvin but fahrenheit must be
converted to celcius and then Kelvin.
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Math Skills #24 Specific Heat
Specific heat is the amount of energy transferred as heat that will raise the
temperature of 1 kg of a substance by 1 K.
SPECIFIC HEAT EQUATION: Q = (specific heat)(mass)(temperature change)
(c) (m) (delta t)
** remember to convert grams to kilograms for all specific heat problems.
Example 1: How much energy is needed to increase the temperature of 755 g of
iron from 283 K to 403 K?
Q = (c)(m)(delta t)
Q = (449 J/kg x K)(0.755 kg)(120 K)
Q = 40,000 J
Example 2: An aluminum baking sheet with a mass of 225 g absorbs 24,000 J of
heat from the oven. If its temperature was initially 25 C, what
will its new temperature be?
delta t = Q / (c)(m)
delta t = (24,000 J / (897 J/kg x K)(0.225 kg)
delta t = 120 C
final t = 120 C + 25 C = 145 C
Example 3: What mass of water is required to absorb 470,000 J of energy from
a car engine while the temperature increases 57 K?
m = Q / (c)(delta t)
m = 470,000 J / (4180 J/kg x K)(57 K)
m = 2.0 kg
Example 4: A vanadium bolt gives up 1,124 J of energy as its temperature
drops 25 K. If the bolt's mass is 93 g, what is its specific
heat?
c = Q / (m)(delta t)
c = 1,124 J / (0.095 kg)(25 K)
c = 480 J/kg x K
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