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Nostrana [21]
3 years ago
15

A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

op-up such that after hitting the bat, the baseball is moving at 53.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.20 ms. What is the average vector force the ball exerts on the bat during their interaction?
Physics
1 answer:
Luda [366]3 years ago
6 0

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

F=ma

F=\dfrac{mv}{t}

F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}

F₁ = 2900 N...........(1)

F=ma

F=\dfrac{mv}{t}

F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

F=F_1+F_2

F=(2900i+(-3493.18)\ N

F=(2900i-3493.18j)\ N

Hence, this is the required solution.

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Answer:

Distance covered to top of the hill was : 1.755 km

Explanation:

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v- 35 / 0.05 = 2

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Formula for distance is product of speed and time

Distance covered = 35.10 * 0.05 = 1.755 km

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The solution for this problem is:
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A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, reaches the ground in 12s. Cal
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