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melisa1 [442]
3 years ago
9

A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu

re p0 and absolute temperature T0. The gas expands isobarically to a volume 2V0, then expands further adiabatically to a final volume of 4V0.
Draw a PV diagram for the squence of process.

Physics
1 answer:
ANTONII [103]3 years ago
8 0

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}

Where:

\gamma = \frac{c_{p}}{c_{v}}

\gamma = 1 + \frac{R}{c_{v}}

Final pressure is:

P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}}  \right)^{\gamma}

P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}

P_{3} = \frac{P_{o}}{2^{\gamma}}

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3 years ago
A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant
german

The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

The work is done to stretch the spring is;

\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J

To learn more about the spring force refer to the link;

brainly.com/question/4291098

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3 0
2 years ago
A plumber is going to put two pipes in a wall, one in front and one in back. The pipes will be touching once they are installed.
beks73 [17]

Answer:

They become the same exact tempature

Explanation:

Since they got connected it should be the same.

8 0
2 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
2 years ago
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OLga [1]

Answer:

Small sparks can lead to huge explosion if they are left unattended.

Explanation:

Small sparks are not harmful but if these sparks happen near some hazardous material or object then it could lead to heavy explosion. If there is some chemical substance near the spark or there are magnetic lines which can explode the spark then these minor sparks could result in heavy disastrous explosion.

4 0
3 years ago
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