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3 years ago

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roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p

ressure p, in atm, as a function of volume V, in liters: p = p0 V-6/5. Here p0 is a constant of units atm⋅L6/5. show answer Incorrect Answer 50% Part (a) Write an expression for the work W done on the gas when the gas is compressed from a volume Vi to a volume Vf.

1 answer:

miskamm [114]3 years ago

3 0

Answer:

W= -2.5 (p₁*0.0012) joules

Explanation:

Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then

In adiabatic compression, work done by mixture during compression is

W= $\int\limits^f_i {p} \, dV$ where f= final volume and i =initial volume, p=pressure

p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= $\int\limits^f_i {K/V^} \, dV$

W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)

W=1/1-γ (p₁Vf-p₀Vi)

W= 1/1-1.40 (p₁*6/5 -p₀*0)

W= -2.5 (p₁*6/5*0.001) changing liters to m³

W= -2.5 (p₁*0.0012) joules

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Any correct ans guess for 30 pts and brainlist

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4 years ago

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Jack and Jill are maneuvering a 3300 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is

Paraphin [41]

Answer:

The workdone by Jack is $W_{jack} = -1050J$

The workdone by Jill is $W_{Jill} = 0J$

The final velocity is $v = 1.36 m/s$

Explanation:

From the question we are given that

The mass of the boat is $m_b = 3300kg$

The initial position of the boat is $P_i = (2 \r i + 0 \r j + 3\r k)m$

The Final position of the boat is $P_f = (4\r i + 0 \r j + 2\r k )\ m$

The Force exerted by Jack $\r F = (-420\r i + 0 \r j + 210\r k) \ N$

The Force exerted by Jill $\r F_{Jill} =(180 \r i + 0\r j + 360\r k)$

Now to obtain the displacement made we are to subtract the final position from the initial position

$\r P = P_f - P_i$

$= (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k )$

$= (2\r i + 0\r j -\r k )m$

Now that we have obtained the displacement we can obtain the Workdone

which is mathematically represented as

$W =\r F * \r P$

The amount of workdone by jack would be

$W_{jack} =\r F * \r P$

$= [(-420\r i +0\r j +210\r k)(2\r i + 0\r j - \r k)]$

$= (-420) (2) + (210)(-1)$

$= -840 - 210$

$=-1050J$

The amount of workdone by Jill would be

$W_{Jill} =\r F * \r P$

$= [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]$

$= (180 )(2) +(360)(-1)$

$=0J$

According to work energy theorem the Workdone is equal to the kinetic energy of the boat

$W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]$

$-1050 = 0.5*3300 [*v^2- (1.1)^2]$

$-1050 = 1650 [v^2 -1.21]$

$0.6363 = v^2 -1.21$

$v^2 = 0.6363+1.21$

$v^2 =1.846$

$v = 1.36\ m/s$

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4 years ago

When gasoline is burned in the cylinder of an engine, it creates a high pressure that pushes the piston. If the pressure is 100.

Marianna [84]

Answer:

$\dot W=2.1074\ hp$

Explanation:

Given:

pressure on the piston, $P=100\ psi=689476\ Pa$

diameter of the piston, $d=3\ in=0.0762\ m$

displacement of the piston, $s=0.05\ m$

time taken in the piston displacement, $t=0.1\ s$

<u>Now, we find the force on the piston:</u>

$F=P.A$

where, A = area upon which pressure acts

$A=\pi.\frac{d^2}{4}$

$A=\pi.\frac{0.0762^2}{4}$

$A=0.00456\ m^2$

$\therefore F=689476\times 0.00456$

$F=3144.2638\ N$

<u>we know that Power is given as:</u>

$\dot W=\frac{F.s}{t}$

$\dot W=\frac{3144.2638\times 0.05}{0.1}$

$\dot W=1572.1319\ W$

$\dot W=2.1074\ hp$

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3 years ago

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Answer:C

Explanation:

Given

weight of object is equal to weight of object

Suppose weight of Planck is W

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balancing Torque

$w\times x-w(\frac{L}{2}-x)=0$

$2x=\frac{L}{2}$

$x=\frac{L}{4}$

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His is a step down transformer since n(primary) is greater than n(seconcary). You relate the input voltage with the ouput voltage with the following equation:

<span>Vout = n2/n1*Vin (n2/n1 is essentially your 'transfer function' that dictates what a specified input would produce) </span>

<span>Solving the equation: </span>

<span>Vin = Vout*n1/n2 = (320V)*(600/300) = 640 V </span>

<span>This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.</span>

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3 years ago