A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the
1 answer:
Answer:
<em>The internal energy change is 330.01 J</em>
Explanation:
Given
the initial volume = 5.75 L
the final volume = 1.23 L
is the external pressure = 1.00 atm
q the heat energy removed = -128 J (since is removed from the system)
expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;
W = -ΔV
W = -1.00 x(1.23 - 5.75)
W = -1.00 x -4.52
W = 4.52 L atm
converting to joules we have
W = 4.52 L atm x 101.33 J/ L atm = 458.01 J
The internal energy change during compression can be calculated thus;
ΔU = q + W
ΔU = -128 J + 458.01 J
ΔU = 330.01 J
Therefore the internal energy change is 330.01 J
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Answer:
Where represent the force for each of the 5 cases
presented on the figure attached.
Explanation:
For this case the figure attached shows the illustration for the problem
We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.
Th formula is given by:
Where G is a constant
represent the mass for the earth
represent the mass for the spaceship
represent the radius between the earth and the spaceship
For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.
Based on this case we can create the following rank:
Where represent the force for each of the 5 cases
presented on the figure attached.
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct