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2 years ago

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Consider a metal rod of uniform temperature. Is it possible that heat spontaneously flow from one end of the rod to the other en

d so that they end up in different temperatures? Why?

1 answer:

raketka [301]2 years ago

4 0

Answer:

No its not possible in that yes heat flow is due to temperature different but with time equilibrium in temperature is reached hence there is no temperature difference at both ends at this time

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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

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