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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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Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge
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Answer:

The charge on each plate is 0.0048 nC

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C = (A×ε)/d

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   = 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

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