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erik [133]
3 years ago
12

A rock is thrown downward from an unknown height above the ground with an initial speed of 10m/s. It strikes the ground 3s later

. Determine the initial
Physics
1 answer:
bazaltina [42]3 years ago
6 0

At the beginning of the 3 sec, its speed is 10 m/s.

Over the course of 3 sec, gravity adds (9.8 x 3) = 29.4 m/s
to its speed, so its speed after 3 sec is (10 + 29.4) = 39.4 m/s.

Its average speed during the 3 sec is

                               (1/2) (10 + 39.4) 

                           =  (1/2) (49.4)  =  24.7 m/s .

In 3 sec, at an average speed of 24.7 m/s,
the rock travels

                     (3) x (24.7) = 74.1 meters.

If there's no air resistance in its path, then it was thrown
from 74.1 meters above the ground.
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Explanation:

It is given that,

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\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}

F_2=\dfrac{F_1}{A_1}\times A_2

F_2=\dfrac{8800}{0.01}\times 0.04

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3 years ago
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Answer:

a = 52s²

Explanation:

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We first need to solve the velocity equation for time (t):

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Plugging in the known values we get,

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Answer:

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We know that for a pipe open at one end and closed at other end the fundamental frequency is given by

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