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Tatiana [17]
4 years ago
5

A person rides a Ferris wheel that turns with constant angular velocity. Her weight is 549.0 N. At the top of the ride her appar

ent weight is 1.500 N different from her true weight. What is her apparent weight at the top of the ride?
Physics
1 answer:
pashok25 [27]4 years ago
8 0

Answer:

Explanation:

Given that the angular velocity is constant.

The actual weight is W = 549N

The ride apparent weight is

W' = 1.5N

Apparent weight at the top of the ride?

Using newton second law of motion

ΣF = m•ar

ar is the radial acceleration

N — W = —m•ar

N = W —m•ar

N = mg —m•ar

N = m(g—ar)

The apparent weight is equal to the normal

W'(top) = N = m(g—ar)

W'(top) = m(g—ar)

W'(top) = mg —m•ar

We know that the actual weight is

W=mg

Also, the apparent weight is

W' = m•ar

Then

W'(top) = Actual weight - apparent

W'(top) = 549—1.5

W'(top) = 547.5 N

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Answer:

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Explanation:

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7 0
3 years ago
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Answer:

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3 years ago
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Answer:

\Delta GPE=-3.92\ 10^9\ J

Explanation:

<u>Gravitational Potential Energy</u>

It's the capacity of an object to do work due to its relative height from a fixed reference point.

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GPE=m.g.h

Where m is the mass of the object, h is its height and g is the acceleration of gravity, g=9.8 m/sec^2

The mass of water is given as

m=8,000,000\ kg

The height above the rocks is h=50 m. Let's compute the GPE

GPE_1=(8,000,000\ kg)(9.8\ m/s^2)(50\ m)

GPE_1=3,920,000,000\ Joule

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The GPE at the bottom, where h=0

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The change of gravitational potential energy is:

\Delta GPE=GPE_2-GPE_1

\boxed{\Delta GPE=-3.92\ 10^9\ J}

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3 years ago
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ΔS= nΔHvap/T,

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3 years ago
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