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Serhud [2]
4 years ago
10

A lever is placed on a fulcrum. A rock is placed on the right end of the lever and a downward (counterclockwise) force is applie

d to the left end of the lever. What measurements would be most effective to help you determine the angular momentum of the system? (Assume the lever itself has negligible mass.)
Physics
1 answer:
DIA [1.3K]4 years ago
7 0

Answer:

measurements that would be most effective to help you determine the angular momentum of the system are

the mass of the rock, the length of the lever on both sides of the fulcrum, and the force applied on the right side of the lever

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A body is moving vertically upwards. It’s velocity changes at a constant rate from 50m/s to 20m/s in 3 sec. What is it’s acceler
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An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
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Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
  • density of aluminium wire, \rho=2.7\times 10^3\ kg.m^{-3}
  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
  • wave speed, v=118\ m.s^{-1}

<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}

\frac{\Delta L}{L} =5.37\times 10^{-4}

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