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natka813 [3]
1 year ago
15

In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add

ition or subtraction of heat.Assume that the gas is initially at pressure p0, volume V0, and temperature T0. In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is γ=Cp/CV=7/5.Note that, unless explicitly stated, the variable γ should not appear in your answers--if needed use the fact that γ=7/5 for an ideal diatomic gas.Part AFind an analytic expression for p(V), the pressure as a function of volume, during the adiabatic expansion.Express the pressure in terms of V and any or all of the given initial values p0, T0, and V0.Correctp(V) = p0(V0V)75Part BAt 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 container during the expansion.Express the work in terms of p0, V0, and V1. Your answer should not depend on temperature.Part CFind ΔU, the change of internal energy of the gas during the adiabatic expansion from volume V0 to volume V1.Express the change of internal energy in terms of p0, V0, and/or V1.Need Part B and C

Physics
1 answer:
rosijanka [135]1 year ago
5 0

Answer:

A: P=Po(Vo/V)^y

B: W=5/2Po(Vo-Vo^7/5 Vi^-2/5)

C: ∆V= 5/2Po[Vo^7/5 Vi^-2/5 -Vo]

Explanation:

Attached is the solution

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you are hiking along a river and see a tall tree on thhe opposite bank. You measure the angle of elevation of the top of the tre
Sidana [21]

Answer:

Explanation:

Let the height of the tree is y and the distance of tree from point B is x.

According to the diagram

tan61 = \frac{y}{x}

x = 0.55 y ..... (1)

tan49.5 = \frac{y}{50+x}

(50 + 0.55y) 1.17 = y ..... from equation (1)

58.5 + 0.644 y = y

0.356 y = 58.5

y = 164.3 ft

3 0
2 years ago
4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat
nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0
2 years ago
a 1250 kg car accelerates from rest to 6.13m/s over a distance of 8.58m calculate the average force of traction
iogann1982 [59]
Use formula, v^2= u^2 + 2as.
The "v" and the "s" of the formula are given.
Since u is 0, just use f=ma.
I hope this helped!
3 0
2 years ago
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor
nalin [4]

Answer:

u_x=38.13\ m/s

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

S=u_yt+\dfrac{1}{2}gt^2

Here u_y=0

S= 57 cm

0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2

t=0.34 s

Now in the horizontal direction

x=u_xt

Here x=13 m

t= 0.34 s

So

13=u_x\times 0.34

u_x=38.13\ m/s

So the initial speed of ball is 38.13 m/s.

7 0
1 year ago
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
2 years ago
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