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Diano4ka-milaya [45]

4 years ago

9

In a cricket match, the 0.16kg ball is bowled with a straight arm. During one particular delivery, the ball leaves the bowler's

hand horizontally, with a speed relative to the ground of 80.8 mph. The bowler's arm is 0.64 m long and the bowler is running at 10.0 mph when the ball is released.
Calculate the centripetal force exerted on the ball in the bowler's hand, in N.

Note: 1 mph = 0.447 ms^-1.

1 answer:

Papessa [141]4 years ago

6 0

Answer:

412.1 N

Explanation:

First of all let's calculate the total speed of the ball as it is released, which is equal to the speed of the ball + the speed of the bowler:

$v=80.8 mph +10.0 mph=90.8 mph$

Now let's convert it into m/s. We know that

$1 mph = 0.447 m/s$

So

$1 mph:0.447 m/s = 90.8 mph:v\\v=\frac{(0.447 m/s)(90.8 mph)}{1 mph}=40.6 m/s$

And now we can calculate the centripetal force, which is given by:

$F=m\frac{v^2}{r}$

where

m = 0.16 kg is the mass of the ball

v = 40.6 m/s is the speed of the ball

r = 0.64 m is the radius of the circular path (equal to the length of the bowler's arm)

Substituting:

$F=(0.16 kg)\frac{(40.6 m/s)^2}{0.64 m}=412.1 N$

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Kinetic Energy:

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$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

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b) Now, the kinetic energy of each nucleus is:

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