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Blababa [14]
3 years ago
9

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s^2 at the rim?
Physics
1 answer:
Anestetic [448]3 years ago
5 0

Answer:

ω = 0.313 rad/s

Explanation:

D = 200 m ⇒  R = D / 2 = (200 m / 2) = 100 m

ac = 9.8 m/s²

The following equation can be used

ac = ω²*R    ⇒    ω = √(ac / R)

then we insert the values in the last formula

ω = √(9.8 m/s² / 100 m) = 0.313 rad/s

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It is 'C', a mixture, because flour and sugar are both a compound.
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Which option gives an object's volume in Sl units?
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Answer:

B.

Explanation:

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Therefore the answer is B.

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Why is the earth sphered but we cant fall off
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3 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
When a current of 0.015 A passes through human body at 240 volts p.d it<br><br>causes​
geniusboy [140]

Explanation:

Given that,

Current, I = 0.015 A

Voltage, V = 240 volts

We need to find the resistance. Using Ohm's law we can find it as follows :

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So, When a current of 0.015 A passes through human body at 240 volts p.d it  causes​  16000 ohms of resistance.

3 0
3 years ago
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