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Tanya [424]
2 years ago
10

An object of mass M is placed on a disk that can rotate. One end of a string is tied to the object while the other end of the st

ring is held by a pole that is located at the center of the disk. Initially, the object is spun in a horizontal circle of radius R at a constant tangential speed of v0, as shown in the figure, such that the tension in the string is T. At a later time, the disk is spun such that the tangential speed of the object is increased to 2v0 . At a later time, the tangential speed of the object is increased to 3v0 . Which of the following diagrams could represent the forces exerted on the object at one of the given speeds? Assume that the force of static friction is negligible. Select two answers.
2T
3T
4T
9T
Physics
2 answers:
Mars2501 [29]2 years ago
7 0

Answer:

Graphic labeled 4t and graphic labeled 9t

polet [3.4K]2 years ago
6 0

Answer: 4t and 9t

Explanation:

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A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.
jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m  </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

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