Science 10 Physics Read pg. 465 - 477 Unit B.

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Science 10 Physics Read pg Unit B

Conversions Review

How to Multiply Fractions
3 x = 3 x = 8 3 x = 9 12 5 21 8 15 9

Unit Conversions (Distance)
E.g km  m 1 km = 1000 m Try: m  m 3750 m 3.75 km x m = 1 km 5850 m 5.85 km x m = 1 km

E.g. 427cm  m 100 cm = 1 m Try: cm  m 427cm x 1 m = 100cm 4.27 m 8.64 cm 865 cm x 1m = 100cm

E.g. 67 mm  m 1000 mm = 1 m Try: mm  m 0.067 m 67 mm x 1 m = 1000 mm 0.765 m 765 mm x 1m = 1000 mm

E.g m  km 1000 m = 1 km 0.580 km 580m x 1 km = 1000 m

Try Unit Conversions (Time)
E.g h  min 1 hr = 60 min 2.75 h x 60 min = 165 min 1 h Try: 42 min  h 42 min x 1h = h 60 min 8

E.g h  s 1 hr = 3600 s 2.10 h x s = 7560 s 1h Try: 3 h  s 3 h x s = s 1 h

Units to know for this unit:
Distance, height = meters (m) Time = seconds (s) Speed, velocity = meters per second (m/s) Acceleration = meters per second2 (m/s2) Work, energy = Joules (J) Force = Newtons (N) Mass = kilograms (kg) Efficiency = percent (%)

Formulas to know for this unit: v = d t
vave = vi + vf 2 a = vf - vi t F = ma W = Fd Ep = mgh Ek = 1/2 mv2 % Efficiency = useful output x 100% total input 11

Rearranging formulas You need to ISOLATE the variable you are trying to solve for What ever mathematical operation you do to one side of the = you need to also do to the other side Ex. v = d Solve for d and solve for t t

Drawing Graphs

Parts of a Graph All graphs should have:
A horizontal axis (or x axis, ALWAYS TIME!) A vertical axis (or y axis) A title Labels on each axis Units for each axis Appropriate scale (numbering on both axis)

Example: Title Distance-Time Graph Label (Units)
*Note: In Physics, time will always be the horizontal axis Scale Label (Units)

Rearrange the following formulas:
vave = vi + vf 1. solve for vi 2 2. solve for vf a = vf - vi 3. solve for vi t solve for t F = ma solve for a W = Fd 6. sove for F Ep = mgh solve for h Ek = 1/2 mv2 8. solve for v % Eff = useful output x 100% 9. solve for total input total input solve for total output

Draw the following graph
Time (s) Distance (m) 0.0 2.0 5.0 4.0 10.0 6.0 15.0 8.0 20.0 25.0 12.0 30.0 14.0 35.0 16.0 40.0

Energy Causes changes in the motion to occur to an object
It can speed objects up, slow them down or change their direction

Uniform Motion Describes a type of movement
It occurs when an object travels in a straight line at a constant speed is difficult to maintain so ….we use AVERAGE SPEED

t  = change d = distance in m or km t = time in s or h
v = d t  = change d = distance in m or km t = time in s or h v = speed in m/s or km/h

Average Speed = distance traveled time v = d  t
v = dfinal – dinitial tfinal – tinitial v = speed (m/s or km/h) d = change in distance (m or km) t = change in time (s or h)

Example 1 A baseball travels 200 m in seconds. What is the average speed of the baseball? ________ ________ Δd Δt v = d = 200 m v = ? t = 1.50 s v = 200m 1.50 s v = 133 m/s

Try: a.) A baseball travels 20.0 m in seconds. Calculate the average speed. d = 20.0 m Δd Δt v = t = 1.50 s v = ? v = 20.0 m 1.50 s v = 13.3 m/s

b. ). If Lance Armstrong bikes 200. 0 m in 10. 0 s,
b.) If Lance Armstrong bikes m in 10.0 s, what is the cyclist’s average speed? Δd Δt v = d = m t = 10.0 s 200.0 m 10.0 s v = v = ? v = 20.0 m/s

c. ). If a train traveled 100 km in 0. 500 hours
c.) If a train traveled 100 km in hours what is its speed in km/h and in m/s? Δd Δt d = 100 km v = 100 km 0.500 h 200 km/h t = h v = v = ? v = 200 km x 1000 m/km 1 h x 60 min/h x 60 sec/min = 200 km/h 3.6 = 55.6 m/s

Example 2 A car travels 1.00 km at a constant speed of 15 m/s. What time is required to cover this distance? ________ ________ Δd Δt v = d = 1.00 km = 1000 m v = 15 m/s t = ? t = d v t = m 15 m/s t = 67 s

Try: a.) How long would it take a car to travel m if its speed was 40.0 m/s? Δd Δt d = 4000 m v = v = 40.0 m/s t = d v = 4000 m 40.0 m/s = 100 s t = ?

b. ). How long would it take a car to travel
b.) How long would it take a car to travel m if its speed was 10.0 m/s? Δd Δt d = 2000 m v = v = 10.0 m/s t = d v = 2000 m 10.0 m/s = 200 s t = ?

c. ). How long would it take a car to travel
c.) How long would it take a car to travel m if its speed was 30.0 m/s? Δd Δt d = 8000 m v = v = 30.0 m/s t = d v = 8000 m 30.0 m/s = 267 s t = ?

Example 3 A motorist travels 406 km in 4 hours and 15 minutes. What is the average speed in km/h and m/s? ________ ________ Δd Δt v = v = 406 km 4.25 h v = km/h d = 406 km v= ? t = 4 hour + 15min 60 min/h = 4.25 h = /3.6 = 26.5 m/s v

Example 4. How far of a distance will a car cover if it travels 2
Example 4. How far of a distance will a car cover if it travels 2.00 m/s for 1.00 min? Δd Δt d = ? v = v = 2.00 m/s d = v t =2.00 x 60.0 = 120 m t = 1.00 min = 60.0 s 32

Distance Time Graphs distance varies directly with time when speed is constant Have the following components: time is the (x-axis) distance is the (y-axis) the slope of the line is the speed of an object

speed describes the rate of motion an object has

Distance-Time Graph  d  t t(s) d(m) 1.0 20 2.0 40 3.0 60 4.0 80 5.0
1.0 20 2.0 40 3.0 60 4.0 80 5.0 100 Distance-Time Graph distance (m)  d  t time (s)

The steepness of the graph is the slope
Example: slope = y2 – y1 x2 – x1 = 80m – 20m 4.0 s- 1.0 s = 20 m/s slope = rise run = y2 – y1 x2 – x1

The steeper the slope the higher the speed.
A B Distance (m) C time (s) Which line shows the greatest speed? The slowest speed?

Try the Following: Make a Distance time graph for the following
Calculate the speed of the boat

Δd Δt 30 m – 10 m 6.0 s – 2.0 s v = = = 5.0 m/s V

Speed- Time Graphs The area under the graph is the distance an object travels The slope of the line gives you information about the speed E.g. A slope of zero (flat line) = uniform motion Upward slope = speed is increasing Downward slope = speed is decreasing

Time t (s) Speed v (m/s) 0.0 5.00 2.0 4.0 6.0 8.0 10.0 Uniform Motion
Speed (m/s) Uniform Motion Time (s)

Calculate the area under the following speed-time graph up to 10.0 s.
Can be used to determine the distance an object travels…… calculate the area under the line Example 1 Calculate the area under the following speed-time graph up to 10.0 s. Time (s) Speed (m/s) 10.0 5.0 10.0 0.0 5.0

Calculate the area under the following speed-time graph up to 10.0 s.
Solution Calculate the area under the following speed-time graph up to 10.0 s. Time (s) Speed (m/s) 10.0 5.0 10.0 area =   w A = (10.0 s)(5.0 m/s) A = 50.0 m 0.0

Calculate distance travelled by an object in 20.0s.
Example 2 Calculate distance travelled by an object in 20.0s. Time (s) Speed (m/s) 20.0 10.0 0.0

Calculate distance travelled by an object in 20.0s.
Solution Calculate distance travelled by an object in 20.0s. Time (s) Speed (m/s) 20.0 10.0 A= 1 b x h 2 A= x 10 A = 100 m 0.0

Try the Following Calculate distance travelled by an object in 40.0s.
30.0 Time (s) Speed (m/s) 20.0 10.0 20.0 0.0 40.0

Solution Calculate distance travelled by an object in 40.0s.
30.0 Time (s) Speed (m/s) 20.0 10.0 20.0 A= 1 b x h 2 A= x 25 A = 500 m 0.0 40.0

1.2 Velocity

Vectors Verses Scalar Scalar quantities: Vector quantities:
involve only magnitude (amount) Vector quantities: involve both magnitude and direction Are drawn using arrows E.g.) Speed = 20 m/s E.g.) Displacement = 20 m [N] Velocity = 20 m/s [N]

when describing a vector, we have two quantities that indicate the direction:
from the -Degrees x or y axis using -Degrees compass directions (N, S, E, or W)

Cartesian Method E.g. 6 m [30°] E.g. 10 m [right] up (90°) left (180°)
down (270°) right (0°) left (180°) 30° 6 m 10 m E.g. 6 m [30°] 40° 2 m 8 m E.g. 10 m [right]

(+) (+) (-) (-) Navigator Method (+) (+) (-) (-) N W E S
angle (°) N of W angle (°) N of E (+) It uses N-S-E-W .. the angle is relative to E (east) or W (west), or a direction at N, S, E or W W E (-) angle (°) S of W angle (°) S of E (-) (-) (-) S

N E S + y axis  y axis + x axis  x axis vector A: vector B:
W + y axis  y axis + x axis  x axis A B 10 km 1.0 m/s 80 65 vector A: vector B: 10 km, 80 E of the y axis 10 km, 80 E of S 1.0 m/s, 26 S of the x axis 1.0 m/s, 26 S of W

+ y axis N 15°N of W  x axis W E + x axis 30° S of E 55°W of S S
C 15°N of W 15º  x axis W E + x axis 30º 55º 30° S of E 55°W of S A B S S  y axis

Distance Verses Displacement
Distance (d) is how far an object travels It is a scalar quantity (magnitude) Displacement (d) Is change in both distance and direction It is a vector quantity (magnitude and direction)

Vectors Direction Vectors in the Same Direction: 10 m 5 m = 15 m
To find the distance: Add them together To find the displacement: Add them together and include the direction 10 m 5 m = 15 m

Example1 A person runs 25 m south and then another 15 m south. (a) What is the distance travelled?    (b) What is the displacement?

Solution A person runs 25 m south and then another 15 m south. (a) What is the distance travelled?    (b) What is the displacement? 25 m 40 m 15 m 40 m South

Vectors in Opposite Directions:
To find the distance: Add them together To find the displacement: take the difference between the two numbers and include the direction 20 m Distance = 25 m Displacement = 15 m E 5 m

a) What is the distance travelled?
Example 2 A plane flies 200 km north and then turns around and comes back 150 km. a) What is the distance travelled? b) What is the displacement?

a) What is the distance travelled?
Solution A plane flies 200 km north and then turns around and comes back 150 km. a) What is the distance travelled? b) What is the displacement? 350 km 150 km 200km 50 km 50 km N

Speed verses Velocity Speed (v): is the rate of an objects motion
It is a scalar quantity The formula is: v= d  t v = speed  d = distance (dfinal – dinitial)  t = time

Velocity ( ) Describes the rate of motion and the direction of the object’s motion It is a vector quantity the formula is: vave = d  t vave = dfinal – dinitial tfinal - tinitial v

Example 3 A student throws a boomerang north. It travels 35.0 m before it turns around and travels 33.0 m back to him. If the total flight of the boomerang took 5.00s, determine the following: a) distance b) displacement c) speed d) velocity

Solution A student throws a boomerang north. It travels 35.0 m before it turns around and travels 33.0 m back to him. If the total flight of the boomerang took 5.00s, determine the following: a) distance d = 35.0 m m = 68.0 m b) displacement d = 35.0 m [N] m [S] = 2.0 m [N] c) speed   v = d/t = 68.0 m/5.00 s = 13.6 m/s

d) velocity   v = d/t = 2.0 m [N]/5.00 s = 0.40 m/s [N]

Example 4 A plane flies south to Edmonton International Airport, which is 465 km from the Fort McMurray Airport. If the flight takes 50.0 minutes what is the average velocity of the plane in km/h and m/s? vave = d  t = 465 km [S] – 0 km 0.8333… h = 558 km/h [S] t = min 60min/h = …. h 558 km/h x = 155 m/s [S] 3600

Example 5 A train travels at 12.0 m/s [E] for 15.0 minutes. What is the displacement of the train? vave = d  t 12.0 m/s [E] = d 900 s = m [E] = 10.8 km [E] t = 15.0 min x 60 s/min = 900 s

When we graph to demonstrate velocity we use a position time graph
time t(s) Position d(m) [E] 0.0 2.0 10.0 4.0 20.0 6.0 30.0 8.0 40.0 50.0 Position d (m) time (s)

Try the Following According to the data below, what is the velocity of the car?

Solution rise Δd slope = = run Δt calculating the average velocity:
= dfinal – dinitial tfinal – tinitial rise run slope = = = 40.0 m – 10.0 m 8.0 s – 2.0 s = +5.0 m/s = 5.0 m/s [E]

Try the Following Plot the following Data.
0.0 to 2.0 s 2.0 to 4.0 s 4.0 to 6.0 s 6.0 to 8.0 s 8.0 to 10.0 s Time interval (s) Plot the following Data. What type of motion is this?

Solution Uniform Motion!

1.3 Acceleration

Acceleration Is a change in velocity (speeding up or slowing down)
The unites = m/s2 Positive (+) Acceleration= velocity Negative(-) Acceleration = velocity (deceleration)

Positive Acceleration
Positive (+) acceleration occurs two ways: If direction is positive (+) and velocity is increasing + direction - direction increasing velocity If direction is negative (-) and velocity is decreasing + direction - direction decreasing velocity

Position - Time Graphs Position (m) [E] Time t (s)
Positive acceleration the slope is increasing

Velocity – Time Velocity (m/s) [E] Time (s)
positive acceleration the slope is increasing the slope gives the acceleration

Negative Acceleration
Negative” (-) acceleration occurs in two ways: If direction is positive (+) and velocity is decreasing + direction - direction decreasing velocity If direction is negative (-) and velocity is increasing + direction - direction increasing velocity

Position (m) [E] Time t (s)
negative acceleration because the slope is decreasing

Time (s) Velocity (m/s) [E]
Negative acceleration since the slope is decreasing The area under gives the distance traveled.

http://videos. howstuffworks

Uniform Motion Accelerated Motion
Time (s, h, etc) Distance (m, km, etc) d-t Graph Time (s, h, etc) Distance (m, km, etc) d-t Graph slope = speed slope of tangent = instantaneous speed

Uniform Motion Accelerated Motion
Time (s, h, etc) Velocity (m/s or km/h) v-t Graph Time (s, h, etc) Velocity (m/s or km/h) v-t Graph area = distance slope = acceleration area = distance

Uniform Motion Accelerated Motion
no a-t Graph a-t Graph Acceleration (m/s2) Time (s, h, etc) area = velocity

acceleration = change in velocity change in time
a = vf – vi t a = v t where: t = change in time in s or h v = velocity in m/s or km/h vi = initial velocity in m/s or km/h vf = final velocity in m/s or km/h a = acceleration in m/s2

Example 1 Rudy falls out of an airplane and after 8.0 s he is travelling at m/s. What is his acceleration? a = vf - vi t = m/s – 0 m/s 8.0 s = 9.8 m/s2 t = 8.0 s vf = m/s vi = 0 m/s

Example 2 The initial speed of a bicycle is 8.0 m/s and it is moving for 6.0 s. If the final speed is 10.0 m/s, what is the acceleration of the bicycle? vi = 8.0 m/s a = vf – vi t a = 10.0 m/s – 8.0 m/s 6.0 s a = 0.33 m/s2 t = 6.0 s vf = 10.0 m/s a = ?

Try the Following What is the acceleration of a car if its speed is increased uniformly from 40 m/s to 70 m/s in 3.0 s?

Solution What is the acceleration of a car if its speed is increased uniformly from 40 m/s to 70 m/s in 3.0 s? a = vf – vi  t a = (70m/s – 40m/s) 3.0s a = 10 m/s2

Example 3 Mike is traveling down Franklin Ave at 50 km/h. He sees Joslin standing at the bus stop and hits the brakes so he can pick her up. How long will it take for him to come to a stop if his acceleration is –5.0 m/s2? vi = 50km/h x 1000 3600 = m/s vf = 0 m/s t = ? a = -5.0 m/s2 a = vf – vi  t -5.0 m/s2 = (0m/s – m/s) t t = m/s -5.0m/s2 t = 2.8 s

Try the Following Jack is traveling down Thickwood at 10 km/h. He sees Amber walking at the side of the road and hits the brakes so he can pick her up. How long will it take for him to come to a stop if his acceleration is –2.0 m/s2?

Solution Jack is traveling down Thickwood at 10 km/h. He sees Amber walking at the side of the road and hits the brakes so he can pick her up. How long will it take for him to come to a stop if his acceleration is –2.0 m/s2? vi = 10km/h x 1000 3600 = m/s vf = 0 m/s t = ? a = -2.0 m/s2 a = vf – vi  t -2.0 m/s2 = (0m/s – m/s) t t = m/s -2.0m/s2 t = 1.38 s

Try the Following Dan is driving down the highway and comes up behind a logging truck traveling 22 m/s. He hits the brakes, and accelerates uniformly at a rate of m/s2 for 5.0 seconds until he reaches the same speed as the truck. What was his initial speed in m/s and km/h?

Solution vi = ? m/s vf = 22 m/s t = 5.0 s a = -2.5 m/s2
Dan is driving down the highway and comes up behind a logging truck traveling 22 m/s. He hits the brakes, and accelerates uniformly at a rate of m/s2 for 5.0 seconds until he reaches the same speed as the truck. What was his initial speed in m/s and km/h? a = vf – vi  t -2.5 m/s2 = (22 m/s – x) 5.0 s (-2.5 m/s2)(5.0 s) = 22 m/s - x -12.5 m/s = 22 m/s - x x = 22m/s m/s x = 34.5 m/s vi= 35 m/s vi = ? m/s vf = 22 m/s t = 5.0 s a = -2.5 m/s2 34.5m/s x 3600 1000 = 124 km/h = 1.2 x102 km/h

Acceleration Due to Gravity
Is 9.8 m/s2… (applies to all objects) It is greater near sea level It is less on the top of a mountain Larger masses have more There is always drag/air resistance… ignore it

When Doing Calculations:
if an object is falling: if an object is going up: a = 9.81 m/s2 a = 9.81 m/s2

Example 1 How fast are you falling after 2.5 s of free fall. Remember a = 9.81 m/s2 vi = 0 m/s a = 9.81 m/s2 t = 2.5 s vf = ? vf = vi + at vf = 0 m/s + (9.81 m/s2)(2.5 s) vf = m/s vf = 25 m/s

Calculating average speed/velocity for constant acceleration
vave = vi + vf 2 vave= average speed vi = initial speed vf= final speed 99

Example 1 A train traveling through the Rocky Mountains, enters the Kicking Horse traveling at 35 m/s. When it reaches the top of the pass 65 minutes later it has slowed down to 15 m/s. What is the average speed of the train? vave = vi + vf 2 Vave = 35 m/s + 15 m/s 2 Vave = 25 m/s

Example 2 A car traveling is travelling up Thickwood at 40 m/s. When it reaches the top of the hill 3 minutes later it has slowed down to 10 m/s. What is the average speed of the car? vave = vi + vf 2 Vave = 40 m/s + 10 m/s 2 Vave = 25 m/s

Try A car traveling is travelling up highway 63 at 80 m/s. When it reaches the Fort McMurray 30 minutes later it has slowed down to 6 m/s. What is the average speed of the car?

Solution A car traveling is travelling up highway 63 at 80 m/s. When it reaches the Fort McMurray 30 minutes later it has slowed down to 6 m/s. What is the average speed of the car? vave = vi + vf 2 Vave = 80 m/s + 6 m/s 2 Vave = 43 m/s

Work and Energy

Force Is any push or pull on an object It is measured in Newtons (N)
Objects remain at rest unless unbalanced forces act upon it

Balanced & Unbalanced Forces
forces are the same size but in the opposite direction cancel each other out. Unbalanced force Forces are in the opposite direction one force is larger than the other

Deceleration (slowing down)
the force is in the opposite direction of the movement energy is transferred from the source of the force to the object that the force is acting upon Accelerating (speeding up) the force is in the same direction as the moving object

To Change the Motion of Objects
A force is needed

F = force (Kg • m/s2) or 1 Newton (N) m = mass (kg)
F = ma F = force (Kg • m/s2) or 1 Newton (N) m = mass (kg) a = acceleration (m/s2) Note: weight is the force due to gravity ( 9.81 m/s2)

Example1 A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it?

Solution Example1 A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it? F = ma F = (1000kg)(2.5 m/s2 ) F = 2500 N F = 2.5 x 103 N

Try the Following A 500 kg car is accelerated at m/s2. What is the force acting on it?

Solution A 500 kg car is accelerated at 4.5 m/s2. What is the force acting on it? F = ma F = (500kg)(4.5 m/s2 ) F = 2250 N F = 2.2 x 103 N

What is the mass of a crate with a weight of 450 N?
Example 2 What is the mass of a crate with a weight of 450 N? F = ma 450 N = m 9.81 m/s2 m = 450 N/9.81 m/s2 m = 45.9 kg F = 450 N m = ? a = 9.81 m/s2

Example 3 What force is needed to accelerate a 500 kg car from rest to 20 m/s in 5.0 s? F = ma F = (500 kg)(4.0 m/s2 ) F = 2000 N F = 2.0 x 103 N F = ? N m = 500 kg a = ? a = vf – vi t a = (20m/s - 0 m/s) 5.0 s a = 4.0 m/s2

Work Occurs when a person lifts a weight, shovels snow or pushes a car
Occurs when a force acts through a distance