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____ [38]
3 years ago
14

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She re

leases the ball at a speed of 25.5 m/s (about 57.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 1060 m/s2 . What is the length of her arm from the pivot point at her shoulder
Physics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

a = \frac{v^2}{R}\\

Given:

a = 1060 m/s^2\\v = 25.5 m/s

 

Plugging in the values in the formula

1060 = \frac{25.5^2}{R}\\R = 0.613 m

so length of his arm is 61.3 cm

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Answer: 13.2 seconds.

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1 foot = 0.305m so,

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S= ut+1/2 at²

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872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

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A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
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Answer:

a) Pb= 200 PA

b).work done= -3600 joules

c).3600joules

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Explanation:

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a). PbVb= PaVa

Pb= (PaVa)/VB

Pb= (600*3)/9

Pb= 1800/9

Pb= 200 PA

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