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Semenov [28]
3 years ago
7

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

container during the expansion. Express the work in terms of p₀, V₀, and V₁. Your answer should not depend on temperature.
Physics
1 answer:
tino4ka555 [31]3 years ago
4 0

Answer:

W=\frac{p_0V_0-p_1V_1}{\gamma-1}

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

pV^{\gamma}=k

where,

p = pressure

V = volume

\gamma=\frac{C_p}{C_V}

k = constant

Therefore, work done by the gas during expansion is,

W=\int\limits^{V_1}_{V_0} {p} \, dV

=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV

=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\

(using pV^{\gamma}=k )

=\frac{p_0V_0-p_1V_1}{\gamma-1}

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A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr
dmitriy555 [2]

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds

The reaction time cannot be more than 1.69515 seconds

4 0
3 years ago
During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61
bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

7 0
3 years ago
Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2
borishaifa [10]

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

5 0
3 years ago
Which sentence states Newton’s third law?
Cerrena [4.2K]

Answer:

A. If two objects collide, each object exerts a force in the same direction as the other.

Explanation:

6 0
3 years ago
An electric immersion heater is put at the bottom of a large tank of water. The water next to the heater becomes warm
djyliett [7]

Answer:

when the water is heated with immersion heater, the water becomes less dense due to which the warm water rises up and the cooler water fills it's space.

4 0
2 years ago
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