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FromTheMoon [43]
3 years ago
5

Tarzan and Jane. Because of your concern that incorrect science is being taught to children when they watch cartoons on TV, you

have joined a committee which is reviewing a new cartoon version of Tarzan. In this episode, Tarzan is on the ground in front of a herd of stampeding elephants. Just in time Jane, who is up in a tall tree, sees him. She grabs a convenient vine and swings towards Tarzan, who has twice her mass, to save him. Luckily, the lowest point of her swing is just where Tarzan is standing. When she reaches him, he grabs her and the vine. They both continue to swing to safety over the elephants up to a height which looks to be about 1/2 that of Jane's original position. To decide if you going to approve this cartoon, calculate the maximum height Tarzan and Jane can swing as a fraction of her initial height.
Physics
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

The maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

Explanation:

Let

m = Mass of Tarzan

M = Mass of Jane

Given

M = 2m

To calculate the maximum height Tarzan and Jane can swing, we make use of the potential energy at their initial and final position.

Reason being that;

At both the initial and final position, velocity is 0, so there's no kinetic energy.

And the potential energy remains the same (i.e constant) at any given point in the system.

Using P.E = mgh.

At initial position, PE1 = mgh

At final position, PE2 = (m + M)gH.

Where h and H represent the initial and final heights.

m + M is the new weight after Jane and Tarzan swing

Equating PE1 to PE2

mgh = (m + M)gH

By substituton (M = 2m)

mgh = (m + 2m)gH

mgh = 3mgH

Make H the subject of the formula

H = mgh/3mg

H = ⅓h

Hence, the maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

From the question, the new height looks to be about ½ that of Jane's original position; i.e. ½h

The calculated height is smaller than what the cartoon is showing;

We can conclude that the cartoon is wrong.

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7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
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Answer:

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5 0
3 years ago
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