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forsale [732]
4 years ago
14

A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi

th a velocity v. Find the change in the temperature T of the gas resulting from the sudden stoppage of the vessel
Physics
1 answer:
Vika [28.1K]4 years ago
6 0

Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

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A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
Drug abuse is characterized by _____.
arlik [135]
There can be mental health effects and psychological dependence
8 0
3 years ago
Read 2 more answers
A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the
Ivenika [448]
64 meters from the base of the cliff.
4 0
3 years ago
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swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the
Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

6 0
3 years ago
A 4.00kg counterweight is attached to a light cord, which is would around a spool. The spool is a uniform solid cylinder of radi
Sergio [31]

Answer:

Explanation:

Given that,

Mass of counterweight m= 4kg

Radius of spool cylinder

R = 8cm = 0.08m

Mass of spool

M = 2kg

The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:

Then we have,

τ(net) = R~ × T~

τ(net) = R~•i × mg•j

τ(net) = Rmg• k

τ(net) = 0.08 ×4 × 9.81

τ(net) = 3.139 Nm •k

The magnitude of the net torque is 3.139Nm

b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:

L~ = R~ × m~v + I~ω

L = mRv + MR v

L = (m + M)Rv

L = (4 + 2) × 0.08

L = 0.48 Kg.m

C. τ =dL/dt

mgR = (M + m)R dv/ dt

mgR = (M + m)R • a

a =mg/(m + M)

a =(4 × 9.81)/(4+2)

a = 6.54 m/s

6 0
3 years ago
Read 2 more answers
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