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Stolb23 [73]
3 years ago
14

Suppose you "dig a hole" 52 feet deep and place a baseball machine in the hole, so that the balls fire out at 96 ft /s and land

on the ground outside the hole. What is the velocity of one of these baseballs when it hits the ground?
Physics
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

76.73 ft/s

Explanation:

Let the final velocity is v.

initial velocity, u = 96 ft/s

g = 32 ft/s²

height, h = 52 feet

use third equation of motion

v² = u² - 2 gh

v² = 96 x 96 - 2 x 32 x 52

v = 76.73 ft/s

Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.

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A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
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Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

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(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

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Hence, this is the required solution.

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