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Chapter 78: Answer Key
The average vertical velocity will be doubled.
The product of these two is displacement.
Since v at the maximum height is zero, one could use v 0 2 + 2 ad.
The vertical displacement d must be four times bigger since v 0 is doubled and squared.
Free-fall problems are not related to mass.
The carts stick together.
The max angle for range is equal between V X and V Y and will result in equal horizontal displacement.
30 degrees and 60 degrees are not the same as 45 degrees.
10 N is the maximum static friction.
The object is not moving fast.
It is possible to relate the height to the initial upward velocity.
The bullet embedded in the wood will cause some unknown amount of mechanical energy to be lost.
Ignore the minus sign when asked about magnitude.
Since the final momentum must be zero, impulse equals change in momentum can be used.
All other factors will only double the height.
The existing 4-resistor is in close proximity to this 4- equivalent resistance.
2 is the equivalent resistance for that section of the circuit.
The 2- equivalent resistance is in a series with the 3-resistor.
The 3-amp current will split when it comes to the branching point.
The current will split evenly since both pathways have equal resistance.
The material is the same.
The wavelength is the same.
The higher wave speed must have caused the wave's frequencies to be raised.
The velocities cannot be determined.
The component of velocity not directed toward or away from the receiver does not contribute to the shift.
0.25 m is 25 cm.
Equal length components would be included in a 45deg vector.
The 45-45-90 triangles with their components would be formed by 135deg, 225deg, and 315deg.
All values will decrease from maximum to minimum.
The reading will be too high.
The rolling object has to use some of the potential energy for rotational and linear energy.
The object is moving slower down the hill.
At the beginning and end of that single oscillation, the velocity is zero.
Net work is not done because there is no change in energy.
The wavelength of the standing wave is fixed by the length of the string.
There will be a cancellation of the Torque caused by gravity on either side.
The planet causes the weight of an object.
The planet has a reaction force on it.
The normal forces exchanged between the bottom surface of the book and the top surface of the table are an action-reaction pair.
The average speed is greater from C to D.
The slope of the line would be horizontal.
The instantaneous velocity at this point is zero.
The instantaneous velocity at these two points does not exist.
Real-world motion is constant.
Students should measure the period repeatedly and take the mean value.
Take a period measurement for 5 to 10 different times and keep the mass and length the same.
The string-mass combo can be pulled out to a certain angle by the protractor.
Take a period measurement for 5 to 10 different mass.
Take a period measurement for 5 to 10 different lengths.
The center of mass is measured from the pivot point.
Pendulums behave like simple harmonic oscillators under the conditions of small angles.
The pendulum needs to be oscillated at larger angles.
One can predict the results to deviate from the expected as the force no longer acts as a simple restorative force.
The mass investigation could be a source of systematic error.
The students may inadvertently be changing the length of the string when they add on different-sized mass.
From the pivot point to the center of mass is the length of the pendulum.
If the students did not compensate for this, they may see an artificial relationship at higher levels in their graph of mass versus period.
The plot's slope would be equal to 4p 2 / g. One could plot T versus L.
The slope is 4 s 2 /m.
The mass will lose speed as it goes back up.
The entire time the mass is inside the loop, it must be going through centripetal acceleration.
The mass will fail to complete the loop if it is moving too slowly.
The centripetal force is supplied by the normal force throughout the trajectory.
A loss of contact between the sliding mass and the track is indicated if the normal force goes to zero.
The top of the loop will have the slowest speed.
This is the most likely place for the mass to fall.
The right-hand side of the equation will be decreased as the speed at the top gets smaller.
The normal force will be zero at some point when mv 2 / r gets small.
The mass can complete the loop if the speed is greater than this.
The mass will fail to complete the loop if the speed is lower.
The top of the loop is where the potential energy is transformed into energy.
The energy is proportional to the number.
The centripetal force is linear with respect to the normal force.
The critical speed is determined by the normal force going to zero.
There is a linear dependence between h and r.
A component of the force due to gravity is either centripetally or centrifugally.
The normal force is acting alone at the points halfway between the top and bottom.
The radial component of the two forces are shown in the drawings.
The centripetal acceleration of the object is not possible because of the twinning components of gravity.
The mechanical energy is being used.
Less energy will be available for the tangential speed needed to make it through the loop since the original gravitational potential energy must now be going into rotational kinetic energy.
The rolling object won't have enough energy to clear the top of the loop since the height was minimally set for the sliding block.
The tension at point B will be higher than in case.
The speed must be increased.
The effect of the additional mass must be considered.
The effective launch velocity will be affected by the rope's effect on mass.
The end of the rope is the same as the object.
At the center of the circle, the rest of the mass is moving at a lower speed.
The net effect of the added mass is to decrease the launch velocity.
The center of mass of an object is tracked by the equations of motion.
The combined rope-plus-object system is being launched from a lower height since the center of mass is lower.
A shorter range for the projectile is caused by a lower launch velocity.
This negative work is done by the component of weight that is tangent.
The tension must do some positive work to maintain the same energy.
The circle can't be perfect because the tension must supply an upward component as well.
Positive and negative work roles must reverse on the downward half of the circular motion.
If there was resistance, there would be losses in our circuit.
When modeling actual resistance inside of components, a "Internal" Resistor is added to the circuit diagram.
If they didn't have infinite resistance, they would create a parallel path.
If they have less than infinite resistance, the addition of voltmeters to the circuit would lower the resistance.
An example of this is to start at the right of the power supply and then trace a loop clockwise through the top parallel resistor and back to our starting point.
The current measured by the ammeter is reduced by this.
If you have one of the two answers correct, the questions will be graded as completely correct (1 point) or incorrect (0 points).
Any correct responses within an individual free-response question will be awarded partial credit.
The guidelines above are based on the scores from the past AP Physics 1 exams.
The actual score ranges are determined by the College Board each year.
The ranges shown are approximations.
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