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ICE Princess25 [194]
3 years ago
8

an athlete whirls a 7.00 kg hammer tied to the end of a 1.3 m chain in a horizontal circle the hammer moves at the rate of 1.0 r

ev/s what is the centripetal acceleration of the hammer
Physics
1 answer:
ElenaW [278]3 years ago
4 0

Answer:

The answer to your question is a = 1.3 m/s²

Explanation:

Centripetal acceleration is the motion of a body that transverse a circular path.

Data

mass = 7 kg

radius = r = 1.3 m

angular rate = w = 1.0 rev/s

centripetal acceleration = a = ?

Formula

a = rw²

Substitution

a = (1.3)(1)²

Simplification and result

a = 1.3 m/s²

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A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
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Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

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V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

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