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BaLLatris [955]
3 years ago
12

A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop​

Physics
1 answer:
hoa [83]3 years ago
5 0

Refer to the attachment

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What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.
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<em>weight = (mass) x (gravity)</em>

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What of the following is the mathematical form of boyles law?
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Read 2 more answers
A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in
melamori03 [73]

Answers:

a) 0.5 m/s^{2}

b) 1.5 N

Explanation:

a) The centripetal acceleration a_{c} of an object moving in a uniform circular motion is given by the following equation:  

a_{c}=\omega^{2} r  

Where:

\omega=1 \frac{rev}{s} is the angular velocity of the ball

r=0.5 m is the radius of the circular motion, which is equal to the length of the string

Then:

a_{c}=(1 \frac{rev}{s})^{2} 0.5 m  

a_{c}=0.5 m/s^{2} This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force F) that is directed towards the center and is equal to the tension (T) in the string:

F=T=m. a_{c}

Where m=3 kg is the mass of the ball

Hence:

T=(3 kg)(0.5 m/s^{2})

T=1.5 N This is the tension in the string

7 0
3 years ago
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