Mastering Physics 3

D, F

Assuming that all cars have equal speeds, which car has the acceleration of the greatest magnitude, and which one has the acceleration of the least magnitude?

the position and velocity and acceleration of the ball

A ball is lodged in a hole in the floor near the outside edge of a merry-go-round that is turning at constant speed. Which kinematic variable or variables change with time, assuming that the position is measured from an origin at the center of the merry-go-round? the position of the ball only the velocity of the ball only the acceleration of the ball only both the position and velocity of the ball the position and velocity and acceleration of the ball

The magnitude of the acceleration of the car at point A is four times that of the car at point E.

Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car at point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E. The magnitude of the acceleration of the car at point A is twice that of the car at point E. The magnitude of the acceleration of the car at point A is the same as that of the car at point E. The magnitude of the acceleration of the car at point A is half that of the car at point E. The magnitude of the acceleration of the car at point A is four times that of the car at point E.

15.0, 26.0, 0, -9.80

Consider a diagram of the ball at time t0.(Figure 2) Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0=y0=0m). However, it is already moving with initial velocity v⃗ 0, whose magnitude is v0=30.0m/s and direction is θ=60.0degrees counterclockwise from the positive x direction. What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."

15.0, 0, 0, -9.80

Consider a diagram of the ball at time t0.(Figure 2) Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0=y0=0m). However, it is already moving with initial velocity v⃗ 0, whose magnitude is v0=30.0m/s and direction is θ=60.0degrees counterclockwise from the positive x direction. What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.

four times as far

Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? the same distance twice as far four times as far half as far by a factor not listed above

v0 = v2 > v1 > 0

How do the speeds v0, v1, and v2 (at times t0, t1, and t2) compare? v0 = v1 = v2 > 0 v0 = v2 > v1 = 0 v0 = v2 > v1 > 0 v0 > v1 > v2 > 0 v0 > v2 > v1 = 0

The acceleration is perpendicular to v⃗ A and directed toward the inside of the track.

Let v⃗ A be the velocity of the car at point A. What can you say about the acceleration of the car at that point? The acceleration is parallel to v⃗ A. The acceleration is perpendicular to v⃗ A and directed toward the inside of the track. The acceleration is perpendicular to v⃗ A and directed toward the outside of the track. The acceleration is neither parallel nor perpendicular to v⃗ A. The acceleration is zero.

The acceleration is perpendicular to v⃗ C and pointed toward the inside of the track.

Let v⃗ C be the velocity of the car at point C. What can you say about the acceleration of the car at that point? The acceleration is parallel to v⃗ C. The acceleration is perpendicular to v⃗ C and pointed toward the inside of the track. The acceleration is perpendicular to v⃗ C and pointed toward the outside of the track. The acceleration is neither parallel nor perpendicular to v⃗ C. The acceleration is zero.

The acceleration is perpendicular to v⃗ D and pointed toward the outside of the track.

Let v⃗ D be the velocity of the car at point D. What can you say about the acceleration of the car at that point? The acceleration is parallel to v⃗ D. The acceleration is perpendicular to v⃗ D and pointed toward the inside of the track. The acceleration is perpendicular to v⃗ D and pointed toward the outside of the track. The acceleration is neither parallel nor perpendicular to v⃗ D. The acceleration is zero.

The acceleration is zero

Let v⃗ F be the velocity of the car at point F. What can you say about the acceleration of the car at that point? The acceleration is parallel to v⃗ F. The acceleration is perpendicular to v⃗ F and pointed toward the inside of the track. The acceleration is perpendicular to v⃗ F and pointed toward the outside of the track. The acceleration is neither parallel nor perpendicular to v⃗ F. The acceleration is zero.

A constant downward acceleration

Now the cannon is pointed straight up and fired. (This procedure is not recommended!) Under the conditions already stated (drag is to be ignored) which of the following correctly describes the acceleration of the ball? A steadily increasing downward acceleration from the moment the cannonball leaves the cannon barrel until it reaches its highest point A steadily decreasing upward acceleration from the moment the cannonball leaves the cannon barrel until it reaches its highest point A constant upward acceleration A constant downward acceleration

Either press the gas or the brake pedal or turn the steering wheel.

You are driving a car at 65 mph. You are traveling north along a straight highway. What could you do to give the car a nonzero acceleration? Press the brake pedal. Turn the steering wheel. Either press the gas or the brake pedal. Either press the gas or the brake pedal or turn the steering wheel.

the position and velocity of the ball

You throw a ball. Air resistance on the ball is negligible. Which of the following functions change with time as the ball flies through the air? only the position of the ball only the velocity of the ball only the acceleration of the ball the position and velocity of the ball the position and the velocity and acceleration of the ball

t1−t0 t2−t1 (t2−t0)/2

The flight time refers to the total amount of time the ball is in the air, from just after it is launched (t0) until just before it lands (t2). Hence the flight time can be calculated as t2−t0, or just t2 in this particular situation since t0=0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. If a second ball were dropped from rest from height ymax, how long would it take to reach the ground? Ignore air resistance. Check all that apply. t0 t1−t0 t2 t2−t1 (t2−t0)/2

A

What is the direction of the acceleration of the object at moment 5?

directions at time step 0, time step 10 = D,F

What is the direction of the acceleration of the object at moments 0 and 10?

downward to the right

Which direction best approximates the direction of a⃗ when the object is at position 1? straight up downward to the left downward to the right straight down

straight up

Which direction best approximates the direction of a⃗ when the object is at position 2? straight up upward to the right straight down downward to the left

straight down

Which direction best approximates the direction of a⃗ when the object is at position 3? upward to the right to the right straight down downward to the right

B

Which of the paths would the cannonball most likely follow if the cannon barrel is horizontal? A B C D

The one fired from 60∘

Which projectile spends more time in the air, the one fired from 30∘ or the one fired from 60∘? The one fired from 30∘ The one fired from 60∘ They both spend the same amount of time in the air

Increase v0 above 30 m/s. Reduce θ from 60 degrees to 45 degrees.

range: R=vxt2=v20sin(2θ)/g Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. Increase v0 above 30 m/s. Reduce v0 below 30 m/s. Reduce θ from 60 degrees to 45 degrees. Reduce θ from 60 degrees to less than 30 degrees. Increase θ from 60 degrees up toward 90 degrees.