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Nata [24]
3 years ago
7

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has

an arm length (hand to shoulder) of 0.60m. We can model its motionas that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing,the gibbon is moving at 3.3m/s. What upward force must a branch provide to support the swinging gibbon?
Physics
1 answer:
Mumz [18]3 years ago
5 0

Upward force provided by the branch is 260 N

<u>Explanation:</u>

Given -

Mass of Gibbon, m = 9.3 kg

Length of the branch, r = 0.6 m

Speed of the movement, v = 3.3 m/s

Upward force, T = ?

The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.

Therefore,

F net = T - mg

F net = ma = mv²/r

Thus,

T = mv²/r + mg

T = m ( v²/r + g)

T = 9.3 [ ( 3.3)² / 0.6 + 9.8]

T = 259.9 N ≈ 260 N

Therefore, upward force provided by the branch is 260 N

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         times
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                   =    (4 pi) x (40.58 x 10¹² m²)

We're only interested in 70% of the total surface area.

                   =   (0.7) x (4 pi) x (40.58 x 10¹²) m²

                   =            3.57 x 10¹⁴  square meters of Earth's surface.

The volume of the water covering that area is

               (the area) times (average depth of 0.95 mile) .

We have to change that 0.95 mile to meters.
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                      (the area) times (0.95 x 1609 meters).

But we're not there yet.  The question isn't asking for the volume.
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We're ready to get the volume in cubic meters.
We're supposed to know that each cubic meter is 1,000 liters,
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Now we're ready to dump all the numbers into the machine and
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This is my answer, and I'm stickin to it.

But ... just like all the other problems you get in high school, the
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I don't know how much effort you put into this problem, but somewhere
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