12 days ago

Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the crate to th

e right with a horizontal force of magnitude 300 N, but friction prevents the crate from sliding. What is the magnitude Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficient of static friction ?s between the crate and the floor be 0.56 and that of kinetic friction, ?k, be 0.47.

Physics

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

It is known that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v are constant on both sides. Therefore we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let the final pressure be P and the temperature $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide equation (2) by equation (3)

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, the left side temperature equals 1.48 times the right side temperature

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