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A cricket ball of mass 148 g is hit at a speed of 185 km/h straight at a nearby fielder, who catches it. If it takes 202 x 10-3 seconds for the fielder's hands to stop the ball, what is the force (assumed to be constant) between the fielder's hands and the ball? (Units: N)

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Step-by-step explanation:

To find the force between the fielder's hands and the ball, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

First, let's convert the velocity of the ball from km/h to m/s:

185 km/h = (185 * 1000) m / (3600) s

= 51.39 m/s

We can use the formula for acceleration:

acceleration = change in velocity / time

The change in velocity can be calculated by subtracting the initial velocity from the final velocity:

change in velocity = final velocity - initial velocity = 0 - 51.39 m/s = -51.39 m/s

Next, let's convert the mass of the ball from grams to kilograms:

148 g = 148 / 1000 kg = 0.148 kg

Now, we can calculate the acceleration of the ball:

acceleration = (change in velocity) / time = (-51.39 m/s) / (202 x 10^-3 s) = -253.91 m/s^2

Finally, we can calculate the force using Newton's second law:

force = mass * acceleration = (0.148 kg) * (-253.91 m/s^2) = -37.66 N

Therefore, the force between the fielder's hands and the ball is approximately 37.66 N.

User Alan Nelson
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