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kirza4 [7]
3 years ago
14

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

rate of 8.05 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.53 rev/s.
A) Which rate of rotation gives the greater speed for the ball?
B) What is the centripetal acceleration of the ball at 8.16 rev/s?
C) What is the centripetal acceleration at 6.35 rev/s?
Physics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

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The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

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Answer:

<h2>135,000 J</h2>

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From the question we have

workdone = 900 × 150

We have the final answer as

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