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Minchanka [31]
3 years ago
14

A ball is thrown horizontally from a height of 19 m and hits the ground with a speed that is five times its initial speed. What

is the initial speed?
Physics
1 answer:
Helen [10]3 years ago
8 0

Answer:

The initial speed of the ball is 3.93 m/s.

Explanation:

It is given that,

A ball is thrown horizontally from a height of 19 m, h = 19 m

Let u is the initial speed of the ball so, its final velocity, v = 5u

Using third equation of motion as :

v^2-u^2=2gh

(5u)^2-u^2=2\times 9.8\times 19

24u^2=372.4

u = 3.93 m/s

So, the initial speed of the ball is 3.93 m/s. Hence, this is the required solution.

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he Constitution made no mention of how many justices should sit on the Supreme Court. Which statement regarding how the number o
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4 years ago
An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is
Alik [6]
We use a fundamental kinematic equation as follows:

V = Vo + g*t. 
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<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>

<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>

<span>d = Vo*t + 0.5g*t^2. </span>
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8 0
3 years ago
What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

8 0
3 years ago
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