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2 years ago

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The PVT behavior of a certain gas is described by the equation of state: P(V − b) = RT where b is a constant. If in addition CV

is constant, show that: (a) U is a function of T only. (b) γ = const. (c) For a mechanically reversible process, P(V − b)γ = const.

1 answer:

alexdok [17]2 years ago

6 0

Answer:

shown in the attachment

Explanation:

The detailed step by step and necessary mathematical application is as shown in the attachment.

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Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.

Viefleur [7K]

Answer:900 feet

Explanation:

Given

Velocity $\left ( V_1\right )=20 mph\approx 29.334 ft/s$

it take 100 feet to stop

Using Equation of motion

$v^2-u^2=2as$

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

$0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )$

$a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2$

When velocity is 60 mph $\approx 88.002 ft/s$

$v^2-u^2=2as$

$0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )$

s=900.08 feet

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2 years ago

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

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

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

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

n and v remain same at both side. so 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 final pressure is P and temp $T_1 {f} and T_2 {f}$

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

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

similarly

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

divide 2 equation by 3rd equation

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

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

thus, temperature on left side is 1.48 times the temperature on right

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2 years ago

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

bazaltina [42]

Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

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2 years ago

What is the initial velocity of the object represented by the graph? ___m/s Graph:

Alex17521 [72]

Answer:

On a velocity-time graph… slope is acceleration. the "y" intercept is the initial velocity. when two curves coincide, the two objects have the same velocity at that time.

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2 years ago

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Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get

olga_2 [115]

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

7 0

2 years ago