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stiv31 [10]
1 year ago
15

Question 5 (1 point)

Physics
1 answer:
Sophie [7]1 year ago
3 0

Main answer- the velocity of 2.2kg ball is 12.566m/s

supporting answer-   In circular motion the velocity is termed as angular velocity

Angular velocity represents the rate of variation of an object's position angle with respect to time.

W =theta/ time

W = length of the arc/ radius of circle × time

body of the answer-2πr/r×t =2π/t

2×π/ 0.5 = 12.566 m/s

final answer- hence the velocity of ball is revolving is 12.566 m/s

to learn more about velocity of a rotating object  link

https://brainly.in/question/46395358

#SPJ1

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A. The force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
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Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

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Gravity exerts a force on the stuntman of ...

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Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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