Physics: An aluminum wire is held between two clamps under zero tension at room temperature. reducing the temperature, which results in a decrease in the wire s equilibrium length, increases the tension in the wire. taking the cross-sectional area of the wire to be 5.75 10-6 m2, the density to be 2.70 103 kg/m3, and young s modulus to be 7.00 1010 n/m2, what strain (δl/l) results in a transverse wave speed of 118 m/s

An aluminum wire is held between two clamps under

Ответ:

manlycool7543

11.01.2022

Physics

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

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, $v=118\ m.s^{-1}$

We have mathematical expression for strain as:

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

Also velocity ofwave in tensed wire:

$v=\sqrt{\frac{T}{\mu} }$ ...................................(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

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

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

0,0(0 оценок)

Ответ:

zamirareece17

20.05.2020

Physics

Color is determined first by frequency. ... As the frequency is increased, the perceived color gradually changes from red to orange to yellow to green to blue to violet. The eye doesn't perceive violet so well. It always seems to look dark compared to other sources at equal intensity.

Explanation:

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question