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3 years ago

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider

1 answer:

3 0

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude. This is all well and
good, but we can't calculate the difference in potential energy
without knowing the mass of the glider.

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Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an i

julia-pushkina [17]

Answer:

1) the entropy generated is Δs= 0.0363 kJ/kg K

2) the minimum theoretical work is w piston = 201.219 kJ/kg

Explanation:

1) From the second law of thermodynamics applied to an ideal gas

ΔS = Cp* ln ( T₂/T₁) - R ln (P₂/P₁)

and also

k= Cp/Cv , Cp-Cv=R → Cp*( 1-1/k) = R → Cp= R/(1-1/k)= k*R/(k-1)

ΔS = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

where R= ideal gas constant , k= adiabatic coefficient of air = 1.4

replacing values (k=1.4)

ΔS = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

ΔS = 1.4/(1.4-1) *8.314 J/mol K * ln( 600K/300K) - 8.314 J/mol K * ln (10 bar/ 1bar)

ΔS = 1.026 J/ mol K

per mass

Δs = ΔS / M

where M= molecular weight of air

Δs = 1.026 J/ mol K / 28.84 gr/mol = 0.0363 J/gr K = 0.0363 kJ/kg K

2) The minimum theoretical work input is carried out under a reversible process. from the second law of thermodynamics

ΔS =∫dQ/T =0 since Q=0→dQ=0

then

0 = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

T₂/T₁ = (P₂/P₁)^[(k-1)/k]

T₂ = T₁ * (P₂/P₁)^[(k-1)/k]

replacing values

T₂ = 300K * ( 10 bar/1 bar)^[0.4/1.4] = 579.2 K

then from the first law of thermodynamics

ΔU= Q - Wgas = Q + Wpiston ,

where ΔU= variation of internal energy , Wgas = work done by the gas to the piston , Wpiston = work done by the piston to the gas

since Q=0

Wpiston = ΔU

for an ideal gas

ΔU= n*Cv*(T final - T initial)

and also

k= Cp/Cv , Cp-Cv=R → Cv*( k-1) = R → Cv= R/(k-1)

then

ΔU= n*R/(k-1)*(T₂ - T₁)

W piston = ΔU = n*R/(k-1)*(T₂ - T₁)

the work per kilogram of air will be

w piston = W piston / m = n/m*R/(k-1)*(T₂ - T₁) = (1/M*) R/(k-1)*(T₂ - T₁) ,

replacing values

w piston = (1/M*) R/(k-1)*(T₂ - T₁) = 1/ (28.84 gr/mol)* 8.314 J/mol K /0.4 * ( 579.2 K - 300 K) = 201.219 J/gr = 201.219 kJ/kg

6 0

3 years ago

If you change the distance between two charged objects, A and B, which of

mars1129 [50]

The factors which are affected will be based on the distance or range between the charged objects A and B

<u>Explanation:</u>

<u>Definition:</u>

The force, F among the two-point size charges A and B are divided by a range r is given by Coulomb formula.

F=9x10^9 AB/r²

<u>Factors:</u>

If the range among them is doubled the force will be decreased to 1/4 of the initial value. And if the range is split, the force will be four times.
If the range among them is tripled, the force will be decreased to 1/9 of the initial value. If the range is decreased to one-third of the new value, the force will rise to Nine times the initial value.

4 0

3 years ago

Read 2 more answers

A boy kicks a football from ground level. The ball takes 3 seconds to reach its maximum height. What is the angle of the initial

ruslelena [56]

<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Time, t = 3 s

Substituting

v = u + at

0 = u + -9.81 x 3

u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

Displacement, s = 15 m

Acceleration, a = 0 m/s²

Time, t = 6 s

Substituting

s = ut + 0.5 at²

15 = u x 6 + 0.5 x 0 x 6²

u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

Initial vertical velocity = rsinθ = 29.43 m/s

Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

$\frac{rsin\theta }{rcos\theta }=\frac{29.43}{2.5}\\\\tan\theta=11.77\\\\\theta=85.14^0$

The angle of the initial velocity with respect to the horizontal is 85.14°

4 0

3 years ago

When the surface of an object is ____ the plane of projection, the surface appears in its true size and shape on the plane of pr

trapecia [35]

"Parallel to" the plane of projection

5 0

3 years ago

George rides his bike to his friend’s house that is 5 kilometers from his house. If he rides his bike at an average speed of 15

Tju [1.3M]

Answer:

The answer is 0.33 H!

Explanation:

8 0

2 years ago