Pre‐Comp Review Questions ‐ 8 th Grade Answers Section 1‐ Units 1. Fill in the missing SI and English Units Measurement SI Unit SI Symbol English Unit English Symbol Time second s second s. Temperature Kelvin K Fahrenheit °F Length metre m Feet ft. Volume (solid) Cubic Metre m 3 Cubic Feet ft 3 Weight (Force) Newton N. Pounds lbs Mass Kilogram kg slug sl 2. Fill in the missing metric prefix and/or numerical value Metric Prefix Symbol Numerical Multiplier Exponential Multiplier (scientific notation) Tera T 1,000,000,000,000 10 12 Giga G 1,000,000,000 10 9 Mega M 1,000,000 10 6 Kilo k 1,000 10 3 Hecto h 100 10 2 Deca da 10 10 1 Base Unit _____________ 1 10 0 Deci d 0.1 10 ‐1 Centi c 0.01 10 ‐2 Milli m 0.001 10 ‐3 Micro µ 0.000001 10 ‐6 Nano n 0.000000001 10 ‐9 Pico p 0.000000000001 10 ‐12 3. Convert 7651 pm to cm 7.651 * 10 ‐7 cm 4. Convert 1.54kg to cg 154000 cg (1.54 * 10 5 ) 5. Convert 7.38 TC to hC 7.38 * 10 10 hC
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Pre‐Comp Review Questions ‐ 8th Grade Answers
Section 1‐ Units
1. Fill in the missing SI and English Units
Measurement SI Unit SI Symbol English Unit English Symbol
Time second s second s.
Temperature Kelvin K Fahrenheit °F
Length metre m Feet ft.
Volume (solid) Cubic Metre m3 Cubic Feet ft3
Weight (Force) Newton N. Pounds lbs
Mass Kilogram kg slug sl
2. Fill in the missing metric prefix and/or numerical value
Metric Prefix Symbol Numerical Multiplier Exponential Multiplier (scientific notation)
Tera T 1,000,000,000,000 1012
Giga G 1,000,000,000 109
Mega M 1,000,000 106
Kilo k 1,000 103
Hecto h 100 102
Deca da 10 101
Base Unit _____________ 1 100
Deci d 0.1 10‐1
Centi c 0.01 10‐2
Milli m 0.001 10‐3
Micro µ 0.000001 10‐6
Nano n 0.000000001 10‐9
Pico p 0.000000000001 10‐12
3. Convert 7651 pm to cm
7.651 * 10‐7 cm
4. Convert 1.54kg to cg
154000 cg (1.54 * 105)
5. Convert 7.38 TC to hC
7.38 * 1010 hC
6. Convert 25cm3 to m3
2.5 * 10‐5 m3
The conversion factor between cm3 to m3 is
1,000,000 cm3 = 1 m3
or
106 cm3 = 1 m3
How do you get that? Start with the conversion from cm to m.
100 cm = 1 m
A cubic meter is a cube 1 metre wide, 1 metre deep and 1 metre tall.
1 m3 = 1 m x 1 m x 1 m
Substitute the conversion factor for meter to centimetre
1 m3 = (100 cm) x (100 cm) x (100 cm)
1 m3 = 1,000,000 cm3 = 106 cm3
7. Convert 30 m/s to mi/hr
67.1 mi/hr
1 mi/hr = 0.44704 m/sec (approx) In other words, 0.44704 m/sec = 1 mi/hr
Section 2‐ Motion
For the following words, write the definition and equation when applicable, and indicate if
the quantity is a vector or a scalar
1. Reference Point When we try to define a location of our object we need another point
from which we will tell the distance or direction or both to pin point the location. The
another point is the reference point.
2. Motion A change in position of an object over time.
3. Distance How much ground an object has covered.
Vector or Scalar? Scalar
Equation:
4. Displacement “How far an object is from a reference point", it is an object's overall
change in position.
Vector or Scalar Vector
Equation: s = sf – si
s = displacement, si = initial position, sf = final position
5. Speed How fast an object is moving.
Vector or Scalar Scalar
Equation:
6. Average Speed Total distance travelled by an object divided by the time to cover that
distance.
Vector or Scalar Scalar
Equation:
7. Velocity The rate at which an object changes its position.
Vector or Scalar Vector
Equation:
8. Instantaneous Velocity Velocity of an object in motion at a specific point in time.
9. Average Velocity Displacement divided by the time.
Equation:
10. Acceleration The rate at which an object changes its velocity.
Vector or Scalar Vector
Equation:
d = displacement
11. Directly Proportional As one amount increases, another amount increases at the same
rate.
Example of 2 quantities that are directly proportional:
How much you earn is directly proportional to how many hours you work
Work more hours, get more pay; in direct proportion.
If you work 2 hours you get paid $40
If you work 3 hours you get paid $60
etc ...
12. Inversely Proportional When one value decreases at the same rate that the other
increases.
Example of 2 quantities that are inversely proportional:
Speed and travel time are Inversely Proportional because the faster we go the shorter
the time.
As speed goes up, travel time goes down.
And as speed goes down, travel time goes up.
13. What are the 3 ways an object can accelerate?
1) increase speed 2) decrease speed 3) change direction
14. Indicate whether the object will be speeding up, slowing down, or not changing speed given the directions of velocity and acceleration
Velocity Acceleration Speeding up/ Slowing down/ No change
+ + Speeding up
+ ‐ Slowing down
+ 0 No change
‐ + Slowing down
‐ ‐ Speeding up
‐ 0 No change
15. Give an example of when an object has a velocity of 0 m/s but is accelerating
When it is changing velocity from 0 m/s to either a positive or negative velocity, at the
point it is at 0 m/s, it is still accelerating. For example: when you throw something up,
it will eventually stop (0 m/s) and then accelerate back down.
Problems
16. An object moves in the y direction with a position as a function of time given by the
equation y(t)=10+6t‐4.9t2.
a. What is the initial position of the object? y(t) = 10 + (6 * 0) – (4.9 * 02)
= 10 m
b. What is the initial velocity of the object?
12
The initial velocity is the coefficient for the middle term (? * t).
6 m/s
c. What is the position of the object after 2s? y(2) = 10 + (6 * 2) – (4.9 * 22)
= 2.4 m
d. What is the acceleration of the object?
Δ
The acceleration is the 2 * the coefficient for the squared term (? * t2).
2 * 4.0 = 9.8m/s2
17. A person hikes 2120 meters east in 25mins, takes a break for lunch for half hour, and
hikes back 1640m west in 20 mins. What was the person’s distance, displacement,
19. A spring with a force constant k=150N/m attached to a wall is hooked onto a box
with a mass of 1.5kg and stretched 30cm from its equilibrium position and held
in place
a. Draw a diagram of the box‐spring system, labeling the springs equilibrium
point (x=0) and its’ stretched position.
b. How much elastic potential energy does the system have when the box is
stretched to 30cm?
X = 30cm = 0.3m
EPE = ½ kx2 = 0.5 * 150 * 0.32 = 6.75J
c. Assume the box is on ground level. How much total mechanical energy does
the block‐spring system have when the block is held in place at 30cm?
TME = EPE only (as at ground level so no GPE and held in place so not moving, so no KE) =
6.75J
d. Use conservation of energy to determine the velocity of the box‐spring
system if the box is released when it reaches the spring’s equilibrium
MEi = MEf
TME = MEi + MEf
KE only (as at ground level so no GPE and at equilibrium, so no EPE) = ½ mv2 = 6.75J
v = √(2EPE/m) = √(2 * 6.75J / 1.5kg) = 3 m/s
e. How far past the equilibrium point will the block compress the spring before
the block finally comes to a stop?
MEi = MEf TME at stop = EPE only (as at ground level so no GPE and stopped, so no KE) = 6.75J
X = 0.3m = 30cm (Same as initial stretch, just the other direction!)
20. A child pulls a wagon (m=20kg) with a force of 25N along a surface with
coefficient of friction 0.1 a distance of 15m. Find the final velocity of the wagon.
Fg = mg = 20kg * 9.8m/s2 = 196N, FN = Fg = 196N
FKE = μKE Fn = 0.1 * 196N = 19.6N
FNET = FA ‐ FKE = 25N – 19.6N = 5.4N
FNET = ma
a = FNET / m = 5.4N / 20kg = 0.27 m/s2
= + 2 Δ
Vf = √( + 2 Δ √(02 + 2*0.27*15) = 2.8 m/s
21. In the diagram, a 650 kg roller coaster car starts from rest at the top of the first hill of its track, which is 24m high, and glides freely to the end of the ride. [Neglect friction.]
a. Where will the car have the most gravitational potential energy? Why? First hill as highest point.
b. Where will the car have the most kinetic energy? Why? End as lowest point.
c. Calculate the total gravitational potential energy of the car and passenger at the top of the first hill. GPE1 = mgh = 650kg * 9.8m/s2 * 24m = 152880J
d. If the 3rd hill is at a height of 12m, calculate the gravitational potential energy at that point GPE3 = ½ * GPE1 = ½ * 152880J = 76440J
e. Find the velocity of the car and passengers on top of the 3rd hill ME1 = ME3 TME = MEi + MEf ME1 = GPE only (As car starts from rest at the top of the first hill.)