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Vedmedyk [2.9K]
3 years ago
10

A tennis ball is dropped from 1.3 m above

Physics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

2.65m/s

Explanation:

Using the equation of motion:

v² = u²+2a∆S where

v is the final velocity

u is the initial velocity

∆S is the change in distance

a is the acceleration

Given

u = 0m/s

a = 9.8m/s²

∆S = 1.3-0.943

∆S = 0.357m

Substituting the given parameters into the formula

v² = 0²+2(9.8)(0.357)

v² = 0+6.9972

v² = 6.9972

v=√6.9972

v = 2.65m/s

Hence the velocity at which it hit the ground is 2.65m/s

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For 3 kg mass: <em>F = \lpha ma,a = \frac{F}{m} = \frac{12N}{3kg} = 4\lpha m / s^{2}</em>

For 6 kg mass: <em>F = \lpha ma,a = \frac{F}{m} = \frac{12N}{6kg} = 2\lpha m / s^{2}</em>


Hope this helped, have a great day!~

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What is the difference between charging by contact and charging by induction in terms of electron transfer.
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