A solid Cute hug sides 0.50m

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

tino4ka555 [31]

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 $pV^{\gamma}=k$ )

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

4 0

3 years ago

What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg.m/s of momentum?

yanalaym [24]

Answer:

The product of the mass and the volume is known as momentum.

According to the law of momentum, it is stated that the two or more bodies remain in a constant state unless an external force is applied in an isolated room.

Momentum depends on the following:-

Mass

Velocity

Momentum = MASS \ X \ VOLUMEMomentum=MASS X VOLUME

\begin{gathered}momentum = 3000kgm/s\\\\mass = 2000kg\\\\velocity =\frac{momentum}{mass}\\\\v= \frac{3000}{2000} \\\\v= 1.5m/s\end{gathered}

momentum=3000kgm/s

mass=2000kg

velocity=

mass

momentum

v=

2000

3000

v=1.5m/s

3 0

3 years ago

Getyour pointsssssssssssssssssssssss

Maurinko [17]

Answer:

thankkkk youuuu soo muchhh

3 0

3 years ago

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A traffic expert wants to estimate the maximum number of cars that can safely travel on a particular road at a given speed. He a

galben [10]

Answer:

Speed s= 19

Explanation:

Take note of the following parameters:

Speed= s,

Number of cars= N,

Number N of cars that can pass a given spot per minute=

N(s)=88s/16+16(s19)2

The principle of differentiation is here:

We let N(s) = N

N = 86s / (17 + 17((s/19)^2))

17N = 86s /(1 + s²/19²)

17N = 361* 86s /(361 + s²)

17N = 31046s /(361 + s²)

Next step;

Differentiate with respect to s

17N = 31046s /(361 + s²)

Remember the quotient rule [u/v]’ = (vu’ - uv’) / v²

Therefore,

u = 31046s =====> du/ds = u’ = 31046

v = (361 + s²) ====>dv/ds = v’ = 2s

17N = 31046s /(361 + s²)

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

The maximum when dN/ds = 0

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

17 * 0 = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

( 31046(361 + s²) - 31046s(2s) ) = 0

31046 [ (361 + s²) - 2s² ] = 0

31046 (361 - s²) = 0

(361 - s²) = 0

(19 - s)(19 + s) = 0

either s = -19 or s = 19, but s > 0

s = 19

4 0

3 years ago

Calculating Displacement from the Area under a Curve Try Use the graph to answer the question What is the total displacement of

Agata [3.3K]

Answer:

175 m

Explanation:

In a velocity vs time graph, displacement is the area under the curve.

We can calculate this as area of a trapezoid:

A = ½ (10 m/s + 60 m/s) (5 s)

A = 175 m

Or, we can split the area into a rectangle and a triangle.

A = (10 m/s) (5 s) + ½ (60 m/s − 10 m/s) (5 s)

A = 175 m

8 0

3 years ago