Answer:
The centripetal acceleration of the stone is 5 m/s²
Explanation:
The length of the string to which the stone is attached, r = 1 m
The speed with which the string is rotated, v = 5 m/s
The centripetal acceleration,
, is given as follows;

Therefore, the centripetal acceleration of the stone found as follows;

The centripetal acceleration of the stone,
= 5 m/s².
Mass always stays the same therefore it can’t be c and d. Volume stays the same because she is not removing anything. Therefore it is a
Answer:
Given,
Frame rate = 25 frames per second
To find,
Time interval between one frame and the next.
Solution,
We can simply solve this numerical problem by using the following process.
Now,
Number of frames = 25
Total time taken to display the given number of frames (ie. 25 frames) = 1 second
To calculate the time interval between one frame and next, we need to divide the time taken to display total number of frames by total number of frames.
So,
Time interval between one frame and next :
= Time taken to display total number of frames / Total frames
= 1/25
= 0.04 second
Hence, time interval between one frame and next is 0.04 second.
Answer:
0.02 sec, 50 Hz
Explanation:
period = 100/5000 = 0.02 sec
frequency = 1/0.02 = 50 Hz
Answer:
The centripetal acceleration at highway speed is greater.
Explanation:
We assume the motion of the car is uniformly accelerated. Let the highway speed be v.
By the equation of motion,


u is the initial velocity, a is acceleration and t is time
Because the car starts from rest, u = 0.

This is the tangential acceleration of the thread of the tire.
The centripetal acceleration is given by

r is the radius of the tire.
Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about

The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.