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3 years ago

If x =(20-4t^2) find average velocity between t=0and t=2sec

Physics

1 answer:

lapo4ka [179]3 years ago

6 0

Answer:

The average velocity is 8 unit per sec

Explanation:

Given as :

The Distance x = 20 - 4t² unit

Change in time Δt = ( 2 - 0 ) s = 2 s

Let the velocity = V unit/s

∴ V = $\frac{\partial (20 - 4t^{2})}{\partial x}$

Or, V = - 8t unit/s

Now velocity at t = 0

V1 = - 8 × 0 = 0 unit/s

And velocity at t = 2 sec

V2 = - 8 × 2 = - 16

So, Average velocity = $\frac{v2 - v1}{2}$ = $\frac{- 16 - 0}{2}$ = -8

Or, $\begin{vmatrix}V\end{vmatrix}$ = 8 unit/sec

Hence The average velocity is 8 unit per sec Answer

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Answer: The change in entropy of the given process is 1324.8 J/K

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Pressure is taken as constant.

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$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

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To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

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$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

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Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

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$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

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Putting values in equation 1, we get:

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If x =(20-4t^2) find average velocity between t=0and t=2sec​

If x =(20-4t^2) find average velocity between t=0and t=2sec