15. An astronaut on the moon has a 110 kg crate and a 230 kg crate. How do the forces required to liftyihe crates straight up on
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Answer:
Power output: W=1426.9MW
Explanation:
The power output of the falls is given mainly by its change in potential energy:
The potential energy for any point can be calculated as:
If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:
If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:
Since the tower base is square with a side length of 125 m,
Therefore,
Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.
The volume of cylinder,
Thus, the mass of the air in the cylinder,
Hence, the mass of the air in the cylinder is this more than the mass of the tower.
Answer:
T₂ = 5646 K
Explanation:
Let's start by finding the power received by the first disc, for this we use Stefan's law
P = σ. A e T⁴
Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1
The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is
I = P / A
Writing this expression for both discs
I₁ A₁ = I₂ A₂
I₂ = I₁ A₁ / A₂
The area of a sphere is
A = 4π r²
I₂ = I₁ (r₁ / r₂)²
r₂ = r₁ ± 5
I₁ = I₂ ( (r₁ ± 5)/r₁)²
.
Let's write the Stefan equation
P / A = σ e T⁴
I = σ e T⁴
This is the intensity that affects the disk, substitute in the intensity equation
σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²
The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white
e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²
T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)
If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term
(1 ±x) n = 1 ± n x
Where x = 5 / r₁ << 1
We replace
T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)
T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)
If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach
T₁ = T₂ {(e₂)}^{1/4} ¼
T<u>₁</u> = T₂ 0.9
5500 = T₂ 0.974
T₂ = 5646 K