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A baseball catcher stops a 93-mph fastball over a distance of 0.1 m. What is the force necessary to stop the 0.14-kg fastball?

Answer :

The force necessary to stop the 0.14 kg fastball is approximately 58.2 Newtons.

To calculate the force necessary to stop the 0.14 kg fastball, we can use the equation:

Force = mass x acceleration. In this case, the mass is 0.14 kg.

To find the acceleration, we need to first convert the velocity of the fastball from mph (miles per hour) to m/s (meters per second).

Given that the fastball is traveling at a speed of 93 mph, we can convert this to m/s by multiplying by a conversion factor: 1 mph = 0.447 m/s.

So, the velocity of the fastball in m/s is 93 mph x 0.447 m/s = 41.571 m/s (rounded to three decimal places).

Now, we can calculate the acceleration using the equation: acceleration = change in velocity / time taken.

In this case, the change in velocity is the initial velocity of the fastball, which is 41.571 m/s, and the time taken is the distance over which the fastball is stopped, which is 0.1 m.

So, the acceleration is 41.571 m/s / 0.1 m = 415.71 m/s^2 (rounded to two decimal places).

Finally, we can find the force by multiplying the mass (0.14 kg) by the acceleration (415.71 m/s^2).

Therefore, the force necessary to stop the 0.14 kg fastball is 0.14 kg x 415.71 m/s^2 = 58.2 N (rounded to two decimal places).

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