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course Phy 201
Problems more complicated then before. I understand rate of change, seems like there is something we have not went over.
q001. On a graph of velocity v (in cm/sec) vs. clock time t (in sec):What are the velocity and clock time corresponding to the point (4, 12)?
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Velocity= 4 cm/sec
Clock time= 12 secs
@& On a graph of v vs. t, the first coordinate is that of the independent variable t, the second coordinate being the dependent variable v.
So for this point t = 4 sec and v = 12 cm / sec.
If you have reversed this order, as many students do, it won't be a serious problem on this question. However do make note of the correct order.
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What are the velocity and clock time corresponding to the point (9, 32)?
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Velocity= 9 cm/sec.
Clock time = 32 sec.
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If these points correspond to the velocity of a ball rolling down an incline, describe as fully as you can what you think happens between the first event (corresponding to the first point of the graph) and the second event (corresponding to the second point of the graph).
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@& At rest the ball is at a stand still and as the ball is released from a finger and rolls down the ramp the velocity increases from the first point at rest to the stopping point.
On this interval the ball speeds up from its initial velocity of 12 cm/sec to its final velocity of 32 cm/sec during a time interval of 5 seconds, lasting from clock time t = 4 sec to t = 9 sec.
If as many students do you read the coordinates in reverse order you will say that the ball speeds up from an initial velocity of 4 cm/sec to its final velocity of 9 cm/sec during a time interval of 20 seconds, lasting from clock time t = 12 sec to t = 32 sec.
Note that either way, on this interval the ball does not start from rest.*@
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What is the change in velocity between these two events?
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9-4=5 cm/sec.
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What is the change in clock time between these two events?
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32+12 = 20 secs.
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What is the average velocity for the interval between these two events?
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9+4=13/2=6.5 cm/s
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What is the average rate at which the velocity changes, with respect to clock time, between these two events?
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6.5 cm/s/5 s
1.3 cm/s^2
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What is the displacement of the object between these two events?
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5000 kg* m/s^2/ 100 kg= 100 kg *a/ 100 kg
50 m/s^2
@& These are not quantities associated with this situation.
vAve = (change in position) / (change in clock time), so
(change in position) = vAve * (change in clock time
You found a correct average velocity, and you know the time interval, so what do you conclude is the change in position?*@
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`q002. A 5 kg mass accelerates at 2 m/s^2. What is the net force acting on the object?
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F_net= m* a
F_net= 5 kg * 2 m/s^2
F-net= 10 kg * m/s^2
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`q003. A net force of 5000 kg m/s^2 acts on a 100 kg mass. What is the acceleration of the mass?
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5,000 kg* m/s^2 / 100 kg = 100 kg *a /100 kg
50 m/s^2
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`q004. A force of 400 Newtons is exerted on an automobile as it is pushed through a distance of 100 meters. How much work is done on the automobile?
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'dW= F * ds
'dW= 400 newtons * 100 m
'dW= 40,000 n*m
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`q005. A certain pendulum requires .5 Newtons of force for every centimeter it is pulled back (recall pulling back the pendulum hanging from the tree limb, using the rubber band).
How much force would be required to pull the pendulum back 5 cm, 10 cm and 15 cm?
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F_net= 0.5 newtons * 5 cm
= 2.5 n * cm
F_net= .5 newtons * 10 cm
= 5 n* cm
F_net= .5 newtons * 15 cm
= 7.5 n * cm
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What is the average force required between pullbacks of 5 cm and 15 cm?
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5 cm +15 cm= 20 cm/2 = 10 cm
@& This would be the average of the two positions, not the average force.*@
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How much work is done between these two positions?
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'dW= .5 newtons * 10 cm
= 5 n * cm
@& Note that 10 cm in this situation is the change in position, 15 cm - 5 cm = 10 cm.
10 cm is also the average of the two positions, so make sure you're thinking in terms of the change in position rather than the average position.*@
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`q006. A system rotates through 12 rotations in 4 seconds, first coming to rest at the end of this interval.
How quickly is it rotating, on the average? (The answer is as obvious as it should seem, but also be sure to interpret this as a rate of change with respect to clock time, and carefully apply the definition of average rate)
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360*12 /4= 1,080 rot./ sec
3 rot/ sec.
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Is it speeding up or slowing down?
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Slowing down because more rotations casue it to slow down
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At what rate is it doing so? (Again attempt to interpret as an average rate of change of an appropriate quantity with respect to clock time).
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12 rot./ 4 sec. = 3 rot./ sec.
@& That's the average rate at which it's rotating, not the average rate at which is it slowing down.
By how much does its rate of rotation change? You know that its average rate of rotation is 3 rot / sec and its final rate is 0, so what are its initial and final rates of rotation?
How much slowing down does it therefore do on this interval, and how long does it take to do so?
At what average rate is it therefore slowing?*@
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`q007. At the initial point of an interval a 7 kg mass is moving at 5 meters / second. By the end of the interval it has gained an additional 200 kg m^2 / s^2 of kinetic energy.
How much kinetic energy does it therefore have at the end of the interval?
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KE= 1/2 (7 kg) (5 m/s)^2
KE= 87.5 m^2/s^2
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Slightly more challenging question: How fast is the mass therefore moving at the end of the interval?
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200 kg * m^2 / s^2/ 87.5 m ^2/ s^2= 2.29 m/s
@& This is the KE at the beginning of the interval.
How much KE is added, and what therefore is the KE at the end of the interval?*@
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`q008. An object begins an interval with a kinetic energy of 20 000 kg m^2 / s^2, and ends the interval with a kinetic energy of 15 000 kg m^2 / s^2.
By how much did its kinetic energy change on this interval? (The answer is as obvious as it might seem, but be careful about whether the answer is postitive or negative.)
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-5,000 kg m^2/ s^2
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More challenging: During this interval, how much work was done on the object by the net force?
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'dW= f *ds
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Note that the change in KE is equal to the work done by the net force.
Check your previous answers on this question and see if you can reason out the work done by the net force.
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Also more challenging: If the average force on the object during this interval had magnitude 200 Newtons, then what was its displacement during this interval?
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200 n/ 7s = 28.57 n/s
@& Not a bad attempt.
However, at this point you would know the work done by the net force, and you know that this work is equal to F_net * `ds.
The 200 N is F_net. Knowing the work done by F_net, and F_net, you can find the desired displacement `ds.*@
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`q009. A mass of 200 grams hangs from one side of a pulley, and another mass from the other side. The gravitational force pulling down on this mass is about 200 000 gram cm / s^2, and the tension in the string pulling it upward is about 180 000 gram cm / s^2.
Pick either upward or downward as the positive direction.
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Positive
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Using + for your positive direction and - for your negative direction, what is the gravitational force on this object?
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+ pulling up
- pulling down
@& The question asked for the gravitational force. You haven't answered that question.
How big is the gravitational force?
In what direction does the gravitational force act?
According to your choice of the positive direction, is the gravitational force therefore positive or negative? Think about whether the gravitational force is in the direction you have chosen as positive. If so, then that force is positive. If not, then it's negative.*@
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Using + for your positive direction and - for your negative direction, what is the tension force on this object?
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+ pulling up
- pulling down
@& How big is the tension force?
In what direction does the tension force act?
According to your choice of the positive direction, is the tension force therefore positive or negative? Think about whether the tension force is in the direction you have chosen as positive. If so, then that force is positive. If not, then it's negative.*@
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Using + for your positive direction and - for your negative direction, what is the net force on this object?
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F_net= 200g * 100 cm/s
= + 20,000 g * cm/s
@& You chose upward as positive.
The net force does have magnitude 20 000 g cm/s^2 (note the s^2), but the gravitational force is greater than the tension force so the net force is downward.
According to your choice the gravitational force would therefore be negative.*@
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Using + for your positive direction and - for your negative direction, what is the acceleration of this object?
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-100 cm/s
+ 100 cm/s
@& You can have only one answer to this question.
You know the net force and the mass. Since
F_net = m a
you can find a.*@
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If the object's displacement during a certain interval is +30 cm, then according to your choice of positive direction, is the displacement upward or downward?
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upward because it was positive
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When you multiply the displacement by the gravitational force, what is your result? Be sure to indicate whether the result is positive or negative.
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-3000 cm^2/s
+300 * -100 cm/s
@& You need to use the given gravitational force, which isn't given, and the + 30 cm displacement, which was just given.*@
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When you multiply the displacement by the tension force, what is your result? Be sure to indicate whether the result is positive or negative.
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+320 * 20,000= 60,000 cm^2/s^2
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When you multiply the displacement by the net force, what is your result? Be sure to indicate whether the result is positive or negative.
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+600,000g * cm/ s^2
+30 * 20,000 g * cm/s^2
@& +30 cm * 20,000 g * cm/s^2 =60 000 g cm^2 / s^2*@
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Does gravity do positive or negative work on this object?
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negative because it pulls down
@& The question is whether the sign of the displacement is the same as or different from the sign of the gravitational force.*@
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Does the tension force do positive or negative work on this object?
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@& The question is whether the sign of the displacement is the same as or different from the sign of the tension force.*@
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Does the net force do positive or negative work on this object?
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negative because overall it is pulling down on the object.
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Does the object speed up or slow down?
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The object slows down because of friction and tension
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How would your answers have changed if you had chosen the opposite direction as positive?
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Then the tension would be the negative direction
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`q010. A pendulum of length 2 meters and mass 3 kg, pulled back a distance | x | from its equilibrium position, experiences a restoring force of magnitude k | x |, where k = 15 kg / s^2 . [Note that for convenience in calculation we are making some approximations here; the actual value of k for this pendulum would actually be closer to 14.7 kg / s^2, and this is so only for values of | x | which are a good bit smaller than the length. These are details we'll worry about later.]
How much force does the pendulum experience when x = .1 meter?
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F_net= m * a
= 3 kg * .1 m
= .3 kg * m
@& .1 m is not an acceleration.
The statement of the problem tells you that the restoring force is of magnitude k | x |, where k = 15 kg / s^2
What is x, and what therefore is the restoring force?*@
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How much force does the pendulum experience when x is .05 meter, .1 meter, .15 meter and .2 meter?
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F_net= 3 kg (.05) =.15 kg * m
= 3 kg (.1 m) = .3 kg * m
= 3 kg (.15 m) = .45 kg * m
= 3 kg (.2 m) = .6 kg * m
@& F_net = mass * acceleration. However .05 meters, for example, is not an acceleration.
You don't know the acceleration, but you were given the rule for the force in the original statement of this problem.*@
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What do you think is the average force between | x | = .05 meter and x = .2 meter?
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F_net = 3 kg (.05m) = .15 kg * m
= 3 kg ( .2m) = .6 kg * m
.15+.6 = .75/2= .375 kg * m
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How much work do you think would be done by this force between | x | = .05 meter and | x | = .2 meter?
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'dW=3 kg (.15m)
= .45 kg * m
@& Your .15 m displacement is correct.
However 3 kg is a mass, not a force.
You need to multiply the average force, not the mass, by the displacement.*@
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How fast would the pendulum have to be going in order for its kinetic energy to equal the result you just obtained for the work?
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.45 kg * m / 2m = .255 kg/ m^2
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If the pendulum moves from position x = .05 meter to x = .2 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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opposite direction because its a gain of +.15
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If the pendulum moves from position x = .20 meter to x = ..05 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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same because it is a loss of -.15
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If the pendulum string was cut, what would be the acceleration of the 1 kg mass?
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If the string is cut the string stops exerting a force and the pendulum is subject only to the force of gravity, which accelerates it downward at 9.8 m/s^2.
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What is the magnitude of the force exerted by gravity on the pendulum's mass? For simplicity of calculation you may use 10 m/s^2 for the acceleration of gravity.
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10 m/s^2 / 3 m = 3.33 s^2
@& The force is equal to mass * acceleration.*@
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When x = .1 meter, what is the horizontal displacement from equilibrium as a percent of the pendulum's length?
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.1/3.33= .003 m/s ^2
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When x = .1 meter, what is the restoring force as a percent of the pendulum's weight?
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3m/.1= 30%
@& 3 m / .1 = 30, not 30%.
30 is 3000 %.*@
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What is the magnitude of the acceleration of the pendulum at the x = .15 meter point?
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.15+0/2= .075 m
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`q011. The force exerted on a mass has magnitude | F | = 15 Newton / meter * x, for 0 <= x <= .25 meter.
Sketch a graph of | F | vs. x. (You might wish to start by making a table of | F | vs. x, for some appropriate values of x between 0 and .25 meter). Note the convention that a graph of y vs. x has y on the vertical axis and x on the horizontal, so that for this graph | F | will be on the vertical axis and x on the horizontal.
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Verify that the points (.06 meter, .9 Newton) and (.16 meter, 2.4 Newtons) lie on your graph.
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The x-axis would be in meters and y-axis would be in newtons. Graphing each by x and y coordinates listed.
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What is the rise between these points?
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2.4 n - .9 n = 1.5 n
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What is the run between these points?
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.16 m - .06 m = .1 m
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What is the average slope between these points?
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1.5 n/.1 m= 15 n/m
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What is the average 'graph altitude' of the 'graph trapezoid' formed by these points?
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.9 + 2.4 = 3.3/2= 1.65 n
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What is the width of the trapezoid?
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width= .1
.16-.06=.1
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What therefore is the area of the trapezoid?
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1.65 n * .1 m =.165 n*m
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What are the graph points corresponding to x = .05 meter and to x = .20 meter?
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(.05,0) (.20, 0)
@& Those points would be on the x axis, not on the graph of force vs. position. At .05 meters the force is not 0; nor is it 0 at .15 meters.
You're good up to this point.
From this point on you are using 0 N as your force. You have the right ideas and the right displacement, but your force information needs to be modified.*@
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What is the area of the 'graph trapezoid' defined by these points?
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0 + 0/2 = 0
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What is the meaning of the altitude of this trapezoid?
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altitude is 0
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What is the meaning of the width of this trapezoid?
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width = .15
.20 - .05 = .15
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What therefore is the meaning of the area of this trapezoid?
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0 * .15 = 0
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`q012. For the rotation data you took in class:
What was the average rate of rotation in the trial where the added masses were at the end of the rotating beam?
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7/2= 3.5 rot./s
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What would then have been the initial rate of rotation (at the instant your finger lost contact with the system)?
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0/2 = 0
@& The system was moving when your finger lost contact.*@
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Assuming that those masses were 14 cm from the center of rotation, how fast were they moving, in cm / second, at that initial instant?
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14/2= 7 cm/s
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Assuming that the masses were each 60 grams, what was the kinetic energy of each mass?
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KE= 1/2 (60 kg) (2 m/s) ^2
= 120 g * m^2/s^2
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"
You
@& You're doing well with the concepts of motion and with many of the force and work ideas.
However check my notes. In some places you have misidentified information, and in some you didn't use the information that was given. Everybody's missing some things, so that isn't a big problem, but be sure to think about my notes and try to identify some of the things you missed.*@