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2 years ago

A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?

Physics

1 answer:

sesenic [268]2 years ago

7 0

Answer:

Speed, u = 29.4 m/s

Explanation:

Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s

Let u is speed with which the ball is thrown up. When the ball falls, v = 0

Using first equation of motion as :

v = u + at

Here, a = -g

So, u = g × t

$u=9.8\times 3$

u = 29.4 m/s

So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.

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Vladimir [108]

Answer:

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Explanation:

We are being told that both stunt professionals are standing from the same height, therefore they will attain the same equivalent speed at the bottom if we are to look at it from the principle of conservation of energy.

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Explanation:

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2 years ago

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

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To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

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According to the values given we have to

$\Upsilon_{steel} = 200*10^9Pa$

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Replacing this values at our previous equation we have,

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

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2 years ago

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Answer:

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2 years ago

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gayaneshka [121]

Answer:

Cis, Trans.

Explanation:

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2 years ago