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Ann [662]
3 years ago
12

A rock climber stands on top of a 50 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s a

part and observes that they cause a single splash. The initial speed of the first stone was 2.4 m/s .
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?
Physics
1 answer:
bagirrra123 [75]3 years ago
3 0
<h2><em>Answer: b) What was the initial speed of the second stone?</em></h2>

Explanation:

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a) 24.43 radians per second

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a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}

\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}

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The distance traveled by the particle at the given time interval is 0.28 m.

<h3>Position of the particle at time, t = 0</h3>

The position of the particle at the given time is calculated as follows;

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<h3>Position of the particle at time, t = 4</h3>

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<h3>Distance traveled by the particle at the given time interval</h3>

d = √[(x₄ - x₀)² + (y₄ - y₀)²]

d =  √[(0.28 - 0)² + (1.98 - 2)²]

d = 0.28 m

Thus, the distance traveled by the particle at the given time interval is 0.28 m.

Learn more about distance here: brainly.com/question/23848540

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