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guapka [62]
4 years ago
6

A batter hits a baseball with a bat. The bat exerts a force on the ball. Does the ball exert a force on the bat?

Physics
2 answers:
pashok25 [27]4 years ago
8 0

Answer:

when batter hit the ball with bat then force is applied by the bat on the ball and same force is applied on the bat by the ball.

This is an example of action reaction law

Explanation:

As we know by newton's III law that every action has equal and opposite reaction force

Which means if a force is applied by an object to another object then the same magnitude force is applied by that object on the first one.

The force will be of same magnitude but opposite in direction

This is known as reaction force.

As per Newton's III law every action has equal and opposite reaction force on it.

So here when batter hit the ball with bat then force is applied by the bat on the ball and same force is applied on the bat by the ball.

This is an example of action reaction law

Mazyrski [523]4 years ago
3 0
Yes, think about the difference of swinging a bat and not hitting a ball. It's fairly easy right? Now, when you hit a ball with the bat, you will feel the bat sting your hands. That's the force the ball is exerting on the bat!
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kati45 [8]

Answer:

The time is 0.90 sec.

Explanation:

Given that,

Height = 4.00 m

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s = ut-\dfrac{1}{2}(-g)t^2

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Put the value into the formula

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Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

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              I = 1.9 * 0.33^2

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The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

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                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

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