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Tatiana [17]
3 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]3 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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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:

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

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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

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