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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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Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

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Explanation:

We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.

By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

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7 0
3 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
a crane lifts four pallets of bricks each of which weigh 5000 N. the crane lifts each pallet a height of 30m. the crane takes 4
love history [14]

Answer:

625 W

Explanation:

Applying

P = W/t.................... Equation 1

Where p = power, W = Work, t = time

But,

W = Force (F) × distance (d)

W = Fd........................ Equation 2

Substitute equation 2 into equation 1

P = Fd/t.................... Equation 3

From the question,

Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds

Substitute these values into equation 3

P = (5000×30)/240

P = 625 Watt

3 0
3 years ago
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