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2 years ago

Atmospheric pressure is greater at the base of a mountain than at

1 answer:

8 0

At the top of the mountain, when he tightens the cap onto the bottole, there is some water and some air inside the bottle. Then he brings the bottle down to the base of the mountain.

The pressure on the outside of the bottle is greater than it was when he put the cap on. If anything could get out of the bottlde, it would. But it can't . . . the cap is on too tight. So all the water and all the air has to stay inside, and anything that can get squished into a smaller space has to get squished into a smaller space.

The water is pretty much unsquishable.

Biut the air in there can be <em>COMPRESSED</em>. The air gets squished into a smaller space, and the bottle wrinkles in slightly.

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What do parentheses mean in a Calc formula? A. Ignore the numbers in parentheses until the end. B. Calculate the numbers in pare

kirill115 [55]

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2 years ago

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What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

The mass of the steel ball is $m = 3.0 \ kg$

The length of arm is $l = 70 \ cm = 0.7 \ m$

The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

$r = \frac{l}{2}$

=> $r = \frac{ 0.7}{2}$

=> $r = 0.35 \ m$

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

$\tau = m_a * g * r + m * g * L$

=> $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=> $\tau = 34.3 \ N\cdot m$

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2 years ago

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.

Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

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3 years ago

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Explanation:

I need a diagram to label these parts.

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3 years ago

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ExtremeBDS [4]

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