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Ostrovityanka [42]
3 years ago
8

A coach is hitting pop flies to the outfielders. if the baseball (m= 145 g) stays in contact with the bat for 0.04 s and leaves

the bat with a speed of 50 m/s, what is the average force acting on the ball?
Physics
1 answer:
vekshin13 years ago
4 0
At the start, the ball is at rest and therefore, u=0 m/s. As it leaves the bat, v= 50 m/s

From equations of motion, v=u+at = at (since u=o)
a=v/t = 50/0.04 = 121250 m/s^2

From Newton's second law,
F=ma = 145/1000 *1250 = 181.25 N
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Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force \times Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = \frac{Impulse}{Time} = \frac{202}{0.244} = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

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Answer:

ax = 2.60m/s^{2}, t = 26.92s

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x (1)

Where (vx)f^{2} is the final velocity, (vx)i^{2} is the initial velocity, ax is the acceleration and  \Lambda x is the distance traveled.

Equation (1) can be rewritten in terms of ax:

(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x

2ax \Lambda x = (vx)f^{2} - (vx)i^{2}

ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}  (2)

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ax = \frac{4900m/s}{2(940m)}

ax = \frac{4900m/s}{1880m}

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So, The acceleration of the plane is 2.60m/s^{2}    

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Rewritten equation (3) in terms of t:

t = \frac{(vx)f - (vx)i}{ax}

t = \frac{70m/s - 0m/s}{2.60m/s^{2}}

t = 26.92s

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

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