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Ivanshal [37]
3 years ago
15

A baseball leaves a bat with a horizontal velocity of 20.0 m/s. When it has left the bat for a time of 0.250 s (Assume air resis

tance is negligible). How far will it have moved horizontally?
Physics
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

The distance is 5 m.

Explanation:

Given that,

Horizontal velocity = 20.0 m/s

Time t = 0.250 s

We need to calculate the horizontal distance

The distance traveled by the baseball with horizontal velocity in time t is given by

Using formula for horizontal distance

d = v\times t

d = 20.0\times0.250

d=5\ m

Hence, The distance is 5 m.

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The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
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Answer:

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Explanation:

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In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo
RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

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T = \sqrt{\frac{m}{k} }     ( 1 )

- The equation for the velocity of a simple harmonic motion

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where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s

and by reeplacing it in ( 2 ):

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B Quartz. Will be your answer of thia
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Read 2 more answers
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