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VMariaS [17]
3 years ago
9

A pendulum consists of a large balanced mass hanging on the end of a long wire. At the point where a 28-kg pendulum has the grea

test potential energy, the angle between the pendulum’s weight and the wire is 89°.What is the wire’s length if the magnitude of the torque exerted by the pendulum’s weight at this position is 1.84 × 104 N•m?
Physics
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

The length of the wire is approximately 67.1 m

Explanation:

The parameters of the pendulum are;

The mass of the pendulum, m = 28 kg

The angle between the pendulum weight and the wire, θ = 89°

The magnitude of the torque exerted by the pendulum's weight, τ = 1.84 × 10⁴ N·m

We have;

Torque, τ = F·L·sinθ = m·g·l·sinθ

Where;

F = The applies force = The weight of the pendulum = m·g

g = The acceleration due to gravity ≈ 9.8 m/s²

l = The length of the wire

Plugging in the values of the variables gives;

1.84 × 10⁴ N·m = 28 kg × 9.8 m/s² × l × sin(89°)

Therefore;

l = 1.84 × 10⁴ N·m/(28 kg × 9.8 m/s² ×  sin(89°)) = 67.0656080029 m ≈ 67.1 m

The length of the wire, l ≈ 67.1 m

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Now, expression for the transfer of heat per unit cell is as follows.

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       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

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Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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