PHYSICS HIGH SCHOOL

A juggler throws a bowling pin straight up with an initial speed of 9.20m/s. How much time elapses before the pin reaches the juggler's hands?

Answers

Answer 1

Answer:

Answer:

1.87 s

Explanation:

Initial speed of throw = 9.20 m/s

Net vertical displacement = 0

The bowling pin would be in free fall i.e. a = 9.8 m/s²

Use the second equation of motion:

s = ut + 0.5at²

0 = (9.20)t-0.5(9.8)(t²)

9.20 = 4.9 t

⇒t = 1.87 s

Thus, the total time of flight, the time elapses before the bowling pin falls in juggler's hand is 1.87 s.

Answer 2

Answer:

A juggler throws a bowling pin straight up with an initial speed, the time that elapses before the pin reaches the juggler's hands is 1.88 s.

Given:
Initial speed, u = 9.2 m/s

The time can be calculated from the second equation of motion. The second equation of motion provides a relation between height, initial speed, acceleration, and time respectively.

From the second equation of motion:

h = ut + at²

When the ball reaches the hands, the distance becomes zero. Therefore, the time is:

0 = 9.2t -0.5 × 9.8t²

9.8t = 18.4

t = 18.4÷ 9.8

t = 1.88 s

Hence, the time that elapses before the pin reaches the juggler's hands is 1.88 s.

To learn more about time, here:

brainly.com/question/82806

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Answers

Answer:

300 Btu per hour

Explanation:

given data

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thick = 12 inch

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outside temperature = 15 degrees F

to find out

How many Btu per hour

solution

we get here How many Btu per hour that is express as

Btu per hour = $(area*temp\ difference)/(r\ value\ 12 + r\ value\ 4)$ ............1

put here value we get

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Answers

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Answers

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Answers

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???????????????????????

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Answers

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Answers

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