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Nov 12 2009 12:59am
Ft 5fg Per Unanswer Question.
You must state what formula you're going to use.

If you do it like this guy it's 7-10 fg / question depending on clarity

Quote

11.
 
How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
  Student Response Value Correct Answer Feedback
A. 40.0 kJ     
B. 1.10 MJ 
C. 20.0 kJ     
D. 0.960 MJ     
E. 10.0 kJ     
Score:  0/1

11.
kinetic energy = 1/2mv^2

vehicle initial
1/2 * 1600 kg * 15 m/s * 15 m/s = 180,000 joules

vehicle final
1/2 * 1600 kg * 40 m/s * 40 m/s = 1,280,000 joules

energy required is final (minus) initial
1,280,000 - 180,000 = 1,100,000 joules (converts to 1.1 MJ)




Quote
1.
 
A rifle bullet is fired at an angle of 30° below the horizontal with an initial velocity of 800 m/s from the top of a cliff 80 m high. How far from the base of the cliff does it strike the level ground below?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 138 m (100%)
Equals 138m (100%)
Score:  0/1

2.
 
Lisa throws a stone horizontally from the roof edge of a 50. meter high dormitory. It hits the ground at a point 60. m from the building. Find the time of flight.
  Student Response Value Correct Answer
1.  not answered  0%  Equals 3.2 s (100%)
Equals 3.2 seconds (100%)
Equals 3.20 s (100%)
Score:  0/1

3.
 
A marble rolls off the edge of a table top with a speed of 2.0 m/s. What is the magnitude of its velocity 0.10 s later?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 2.2 m/s (100%)
Score:  0/1

4.
 
A mortar shell is launched with a velocity of 100 m/s at an angle of 30.0° above horizontal from a point on a cliff 50.0 m above a level plain below. How far from the base of the cliff does the mortar shell strike the ground?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 963 m (100%)
Equals 963m (100%)
Score:  0/1

5.
 
A child drops a toy at rest from a point 4.0 m above the ground at the same instant her friend throws a ball upward at 6 m/s from a point 1.0 m above the ground. At what distance above the ground do the ball and the toy cross paths?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 2.78 m (100%)
Score:  0/1

6.
 
In attempting to jump up a waterfall, a salmon leaves the water 2.0 m from the base of the waterfall. With what minimum speed must it leave the water in order just to make it up a waterfall 1.6 m high?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 8.96 m/s (100%)
Score:  0/1

7.
 
A load of steel of mass 6000 kg rests on the flatbed of a truck. It is held in place by metal brackets that can exert a maximum horizontal force of 8000 N. When the truck is traveling 20.0 m/s, what is the minimum stopping distance if the load is not to slide forward into the cab?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 150 m (100%)
Equals 150m (100%)
Score:  0/1

8.
 
What is the weight of a 14.0 kg pumpkin in Newtons?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 137 Newtons (100%)
Equals 137N (100%)
Equals 137 N (100%)
Score:  0/1

9.
 
Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high did it travel?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 32.8 m (100%)
Equals 32.8m (100%)
Score:  0/1

10.
 
Suppose David Ortiz hits a home run which travels 361. feet (110. m). Assuming it left the bat at 50.°, how fast was it hit?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 33.1 m/s (100%)
Equals 33.1m/s (100%)
Score:  0/1

11.
 
A family goes on vacation for one week, but forgets to turn off an electric fan that consumes electricity at the rate of 200 watts. If the cost of electricity is $0.120/kW-h, how much does it cost to run the fan?
  Student Response Value Correct Answer
1.  not answered  0%  Equals $4.03 (100%)
Equals 4 dollars (100%)
Equals $4 (100%)
Score:  0/1

12.
 
FIGURE 5-8


NOTE: The following question(s) refer(s) to the Cyclone, the famous roller coaster ride at Coney Island, shown in the sketch. Assume no friction.


Refer to Fig. 5-8. If the roller coaster leaves point Q from rest, how fast is it traveling at point R?

  Student Response Value Correct Answer Feedback
A. 22 m/s     
B. 0.49 km/s     
C. 51 m/s     
D. 31 m/s     
E. 0.98 km/s     
Score:  0/1

13.
 
A skier, of mass 60. kg, pushes off the top of a frictionless hill with an initial speed of 4.0 m/s. How fast will she be moving after dropping 10. m in elevation?
  Student Response Value Correct Answer Feedback
A. 49. m/s     
B. 15. m/s     
C. 0.15 km/s     
D. 10. m/s     
E. 0.20 km/s     
Score:  0/1

14.
 
You lift a 10. lb physics book up in the air a distance of 1 ft, at a constant velocity of 0.5 ft/s. The work done by gravity is
  Student Response Value Correct Answer Feedback
A. +10 ft∙lb.     
B. +5 ft∙lb.     
C. -5 ft∙lb.     
D. -10 ft∙lb.     
E. zero.     
Score:  0/1

15.
 
A container of water is lifted vertically 3.0 m, then returned to its original position. If the total weight is 30. N, how much work was done?
  Student Response Value Correct Answer Feedback
A. 45 J     
B. 0.90 kJ     
C. 90 J     
D. No work was done.     
E. 0.18 kJ     
Score:  0/1

16.
 
Matthew pulls his little sister Sarah in a sled on an icy surface (assume no friction), with a force of 60.0 Newtons at an angle of 37.0° upward from the horizontal. If he pulls her a distance of 12.0 m, the work he does is
  Student Response Value Correct Answer Feedback
A. 720 J     
B. 433 J     
C. 185 J     
D. 575 J     
E. 0.333 kJ     
Score:  0/1

17.
 
FIGURE 5-5



The force that a squirrel exerts on a nut it has found is observed over a 10. second interval, as shown in Fig. 5-5. How much work did the squirrel do during that 10. s?

  Student Response Value Correct Answer Feedback
A. 22. J     
B. zero     
C. 25. J     
D. 12.5 J     
E. 50. J     
Score:  0/1

18.
 
Is more work required to increase a car's speed from rest to 30 mph, or from 50 mph to 60 mph?
  Student Response Value Correct Answer
1.  not answered  0%  Equals 50 to 60 mph (100%)
Equals 50mph to 60 mph (100%)
Equals 50 to 60 (100%)
Equals 50-60 (100%)
Score:  0/1

19.
 
An object of mass 10.0 kg is initially at rest. A 100 N force causes it to move horizontally through a distance of 6.00 m. What is the change in the kinetic energy of this object?
  Student Response Value Correct Answer Feedback
A. 600 J     
B. 1000 J     
C. 60.0 J     
D. 200 J     
E. 0 J     
Score:  0/1

20.
 
A 4.0 kg box of fruit slides 8.0 m down a ramp, inclined at 30.° from the horizontal. If the box slides at a constant velocity of 5.0 m/s, the work done by gravity is
  Student Response Value Correct Answer Feedback
A. -0.16 kJ.     
B. zero.     
C. +78. J.     
D. -78. J.     
E. 0.16 kJ.     
Score:  0/1

21.
 
A ball, of mass 100. g, is dropped from a height of 12. m. Its momentum when it strikes the ground is
  Student Response Value Correct Answer Feedback
A. 3.3 kg∙m/s.     
B. 1.5 kg∙m/s.     
C. 4.8 kg∙m/s.     
D. 1.8 kg∙m/s.     
E. 2.4 kg∙m/s.     
Score:  0/1

22.
 
Daisy (50. kg mass) skates on ice at 4.0 m/s to greet her friend (70. kg mass), who is standing still, with open arms. As they collide, while holding each other, with what speed do they both move off together?
  Student Response Value Correct Answer Feedback
A. 5.0 m/s     
B. zero     
C. 1.7 m/s     
D. 2.5 m/s     
Score:  0/1

23.
 
Car A (mass = 1000 kg) moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with car B (mass = 200 kg) in a completely inelastic collision. What is the speed of the car immediately after the collision?
  Student Response Value Correct Answer Feedback
A. 3.0 m/s     
B. 6.0 m/s     
C. 5.0 m/s     
D. 4.0 m/s     
E. zero     
Score:  0/1

24.
 
A handball of mass 0.10 kg, traveling horizontally at 25. m/s, strikes a wall and rebounds at 19. m/s. What is the change in the momentum of the ball?
  Student Response Value Correct Answer Feedback
A. 1.8 N-s     
B. 4.4 N-s     
C. 5.4 N-s     
D. 72 N-s     
E. 1.2 N-s     
Score:  0/1

25.
 
A 60. kg person walks along a 100. kg log at the rate of 0.8 m/s (with respect to the log). With what speed (in m/s) does the log move, with respect to the shore? ENTER NUMBER ONLY! NO UNITS.
  Student Response Value Correct Answer
1.  not answered  0%  Equals 0.3 (100%)
Score:  0/1

26.
 
When a light beach ball rolling with a speed of 6 m/s collides with a heavy exercise ball at rest, the beach ball's speed after the collision will be approximately
  Student Response Value Correct Answer Feedback
A. 6 m/s.     
B. 0.     
C. 12 m/s.     
D. 18 m/s.     
E. 3 m/s.     
Score:  0/1

27.
 
Car A (mass = 1000 kg) moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with car B (mass = 200 kg) in a completely inelastic collision. What is the momentum after the collision if car B was initially at rest?
  Student Response Value Correct Answer Feedback
A. 2.0 KN-s to the right     
B. 8.0 KN-s to the left     
C. 6.0 KN-s to the right     
D. 10. KN-s to the right     
E. zero     
Score:  0/1

28.
 
A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum?
  Student Response Value Correct Answer Feedback
A. 0.0280 kg∙m/s downwards     
B. 0.308 kg∙m/s upwards     
C. 0 kg∙m/s     
D. 0.308 kg∙m/s downwards     
E. 0.0280 kg∙m/s upwards     
Score:  0/1

29.
 
An empty coal-car (mass 20,000.kg) of a train coasts along at 10.m/s. An unfortunate 3000. kg elephant falls from a bridge and drops vertically into the car. Determine the speed of the car immediately after the elephant is added to its contents, in meters per second. ENTER NUMBER ONLY! NO UNITS.
  Student Response Value Correct Answer
1.  not answered  0%  Equals 8.7 (100%)
Score:  0/1

30.
 
A 3. kg ball of putty moving at 1 m/s collides and sticks to a 2 kg ball initially at rest. The putty and ball then move with a momentum of
  Student Response Value Correct Answer Feedback
A. 4 N-s.     
B. 1 N-s.     
C. 0 N-s.     
D. 2 N-s.     
E. 3 N-s.     
Score:  0/1

31.
 
An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pipe diameter is  is  What is the fluid speed at a location where the diameter has narrowed to
HINT: use the "flow rate" equation and don't forget that diameter and area are not the same thing.


  Student Response Value Correct Answer Feedback
A. 5.12 m/s     
B. 2.56 m/s     
C. 0.32 m/s     
D. 1.28 m/s     
E. 0.64 m/s     
Score:  0/1

32.
 
A plastic block of dimensions 2.0 cm × 3.0 cm × 4.0 cm has a mass of 30. grams. What is its density?
  Student Response Value Correct Answer Feedback
A. 0.80 g/cm3     
B. 2.3 g/cm3     
C. 1.6 g/cm3     
D. 1.2 g/cm3     
E. 1.3 g/cm3     
Score:  0/1

33.
 
A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area 40 cm2, how much weight can the large piston support?
HINT: see page 322.

  Student Response Value Correct Answer Feedback
A. 10000 N     
B. 25 N     
C. 500 N     
D. 1000 N     
E. 40000 N     
Score:  0/1

34.
 
A ship that weighs 10,000 tons (mass 10 million kg) must displace what volume of fresh water in order to float?
  Student Response Value Correct Answer Feedback
A. 104 m3     
B. 1010 m3     
C. 108 m3     
D. 107 m3     
E. 1012 m3     
Score:  0/1

35.
 
Suppose that an 80. kg person walking on crutches supports all his weight on the two crutch tips, each of which is circular with a diameter of 1.0 inch. What pressure is exerted on the floor? Express the answer in psi. HINT: this is a simple pressure problem. Area is easy to find, F is easy to find, calculate p.
  Student Response Value Correct Answer Feedback
A. 14. psi     
B. 1.1 × 102 psi     
C. 2.2 × 102 psi     
D. 56 psi     
E. 1.4 × 102 psi     
Score:  0/1

36.
 
Substance A has a density of 3 g/cm3 and substance B has a density of 4 g/cm3. In order to obtain equal masses of these two substances, the ratio of the volume of A to the volume of B will be equal to: __________
HINT: It's either 3/4 or 4/3.  Figure out whether there's a direct or inverse relationship between volume and density, and then you'll know which is the correct answer.


  Student Response Value Correct Answer Feedback
A. 1 : 4     
B. 4 : 3     
C. 3 : 4     
D.  : 1     
E. 1 : 3     
Score:  0/1

37.
 
An iron barge 25. m long and 4.0 m wide has a mass of 100,000.kg. What minimum depth of water (in meters) is needed to float it? Enter the number only. HINT: Refer to section 9.3 of the text, and set the buoyant force (up) equal to the weight of the barge (down). Use that to figure out the volume of the water displaced by the barge. Solve for the depth of that water.
  Student Response Value Correct Answer
1.  not answered  0%  Equals 1 (100%)
Score:  0/1

38.
 
A 3.00-m long tube with circular crosssection tapers from a diameter of 10.0 cm at one end down to 4.00 cm at the other end. If an incompressible fluid flows irrotationally through the tube at a speed of 2.00 m/s at the 10.0 cm end, what is the speed of the flow at the 4.00 cm diameter end?
HINT: use the "flow rate" equation and don't forget that diameter and area are not the same thing.

  Student Response Value Correct Answer Feedback
A. 12.5 m/s     
B. 0.800 m/s     
C. 6.25 m/s     
D. 1.60 m/s     
E. 2.00 m/s     
Score:  0/1

39.
 
A 13000 N vehicle is to be lifted by a 25 cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter piston to accomplish this?
HINT:  see page 322.

  Student Response Value Correct Answer Feedback
A. 520 N     
B. 2600 N     
C. 5200 N     
D. 13000 N     
E. 260 N     
Score:  0/1

40.
 
A cylindrical rod of length 12. cm and diameter 2.0 cm will just barely float in water. What is its mass in grams? Enter number only, to 1 decimal place. HINT: set the volume of the rod equal to the volume of water displaced by it, and use the formula on page 329.
  Student Response Value Correct Answer
1.  not answered  0%  Equals 37.7 (100%)
Score:  0/1
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Nov 12 2009 01:16am
8.
Newtons = kg * graviational constant

on Earth gravitational constant is 9.8 m/s

So 1kg = 9.8N (on Earth)

14 kg * 9.8 = 137.2N (rounds to 137)
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Nov 12 2009 01:56am
35. (I don't think I solved this one they way they wanted, hope this does not matter).

area of 1 crutch on floor:
area of a circle = pi * radius * radius
diameter is 1 so radius is 0.5
area = 3.14 * 0.5 * 0.5 = 0.785 square inchs

there are 2 crutches:
area = 2 * 0.785 = 1.57 square inches

weight of fatty on crutches:
pounds = kg / 0.45359237
80 kg / 0.45359237 = 176.36981 pounds

PSI = pounds/square inches
176.37 pounds / 1.57 square inches = 112.3375 psi
Answer B is 1.1 X 102 psi = 112.2 psi
So my answer would round to this answer.
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Nov 12 2009 02:08am
11.
if the bulb was 1000 watts, it would consume 1 kW-h, per hour

so a 200 watt bulb uses 0.2 kh-h, per hour.

It was left on for:
24 (hours) * 7 (days) = 168 hours

Total power consumed:
0.2kW-h * 168 hours = 33.6 kW-h

Price of power:
$0.12 * 33.6 kW-h = $4.032 (rounds to 4 bucks)
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Nov 12 2009 02:08am
11.
Fan consumed 200w/h or converting to Kw .2 Kw-h (divide by 1000)
Multiply by price of electricity and fan costs $.024/h to run (.2 * .12)
Multiply by 24hours in a day then by 7 days in a week to get cost / week .024 * 24 * 7 = 4.03

12.
Requires picture, but you probably want to use potential energy, its easy without friction use starting height's pot energy and ending height's pot energy for change in velocity
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Nov 12 2009 02:30am
1.

vector velocity of bullet = 800m/s
break vector into x and y components
x component of velocity = 800cos30 in the positive direction
= 692.8 m/s
y component of velocity = 800sin30 in the negative direction (down)
= 400 m/s

Bullet travelling 400m/s downwards, cliff height is 80m high, so find time bullet takes to hit ground
t = d/v
t = 80/400
t= 0.2 seconds

bullet takes 0.5 seconds to hit ground, so find how far travelled in y direction (away from base of cliff)
use y component of velocity = 692.8 m/s
d = vt
d = 692.8 x 0.2
d = 138.5 m
round to 138

formulas for velocity vectors = x component = vcos(theta) y component = vsin(theta)

this question does not take into account acceleration from what i can tell because the time interval is so small, but if you want me to add that in lmk (will be very small value)

This post was edited by graemega on Nov 12 2009 02:33am
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Nov 12 2009 02:45am
2.

t= unknown
vi = unknown (velocity initial)
d(y) = y distance = 50m
d(x) = x distance = 60m

t= d(x)/vi(x)
t = 60/vi(x)

d(y) = vi(y)t + 1/2at^2
substitue t for 60/vi(x)
50 = vi(y)(60/vi(x)) + 1/2(9.81)((60/vi(x))^2) after substituting a = 9.81 and d(y) in
note that the velocity in y direction is initially zero, so vi(y) = 0
50 = 17658/vi^2
50vi(x)^2 = 17568
vi(x)^2 = 353.16
vi(x) = 18.8 m/s
put this back into original equation
t = 60/vi(x)
t= 60/18.8
t = 3.2 seconds
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Nov 12 2009 05:19am
21.
Velocity when it hit the ground:
VelocitySquared = VelocityInitialSquared + 2 * acceleration * (PositionFinal - PositionInitial)
VelocityInitial is 0
PositionFinal is 0 (the ground)
PositionInitial is 12 meters
acceleration is the gravitational constant (but it is NEGATIVE because the ball will be going from position 12 to position 0).

V^2 = 0^2 + 2(-9.8)(0-12)
v^2 = 2(-9.8)(-12)
v^2= 235.2
v = 15.336232 m/s


Momentum = mass * velocity (p=mv):
mass of ball in kg is 0.1 kg
p = .1 * 15.336232 = 1.534 kgm/s (rounds to 1.5 kg*m/s)
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Nov 12 2009 05:32am
22.
Initial momentum of Daisy:
p=mv
p= 50kg * 4 m/s = 200kgm/s

Initial momentum of Daisy's fat friend:
p=mv
p= 70kg * 0 m/s = 0 kgm/s

momentum of system = Daisy momentum + chubby's momentum
200kgm/s + 0 kgm/s = 200kgm/s
This is the total momentum in the starting condition and the final condition must match this.

Since they are going to grab on to each other we will consider them as a single object:
p=mv
200kgm/s = 120 kg * Z m/s (divide both sides by 120)
Z = 200/120 = 1.666666 meters per second (rounds to 1.7 meters per second)
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Nov 12 2009 08:03am
23.
p=mv

momentum before collision:
1000 kg * 6 m/s = 6000 kgm/s (right)

momentum after collision:
1200 kg * X m/s = 60000 kgm/s (right)
divide both sides by 1200
X = 5 m/s




24.
p=mv

Approaching wall (we'll call it east):
0.10 kg * 25 m/s = 2.5 kgm/s (east)

Rebounding off wall (we'll call it west):
0.10 kg * 19 m/s = 1.9 kgm/s (west)

To go from 2.5 kgm/s (east) to 1.9 kgm/s (west) requires a change of
2.5 + 1.9 = 4.4 kgm/s (kgm/s is a Newton, so that is the answer)

This post was edited by Azrad on Nov 12 2009 08:06am
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