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Elza [17]
3 years ago
12

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

udely 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 252 m in diameter, what angular velocity (in rad/s) would produce an "artificial gravity" of 9.80 m/s2 at the rim?
Physics
1 answer:
devlian [24]3 years ago
4 0

Answer:

0.278 rad/s

Explanation:

d = Diameter of space station = 252 m

r = Radius = \frac{d}{2}=\frac{252}{2}\ m

g = Acceleration due to gravity = 9.80 m/s²

\omega = angular velocity

Here the centripetal acceleration and the acceleration due to gravity will balance each other

Centripetal acceleration is given by

a_c=r\omega^2

a_c=g\\\Rightarrow r\omega^2=g\\\Rightarrow \frac{252}{2}\omega^2=9.8\\\Rightarrow \omega=\sqrt{\frac{9.8\times 2}{252}}\\\Rightarrow \omega=0.278\ rad/s

The angular velocity required would be 0.278 rad/s

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152642.2

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3 0
3 years ago
a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration th
vekshin1

the average acceleration of the bullet through the board   -55657×10⁵ m/s²

acceleration-  Rate of change of velocity with respect to time.

S = 0.10 mm = 10⁻⁵ m  ( distance)          [∵ 1mm = 10⁻³]

u =  480 m/s (initial velocity)

v = 345 m/s (final velocity)

As we know the 3rd equation of motion.  

v² = u² + 2aS

a = ? ( acceleration)

using these values in equation we get

(345)² = (480)² + 2×a× 10⁻⁵

a = (345)² -  (480)² / 2× 10⁻⁵

a =  -55657×10⁵ m/s²

The negative sign shows that the direction of acceleration is opposite of velocity thus bullet is slowing down.

the average acceleration of the bullet through the board  -55657×10⁵ m/s²

The given question is incomplete. The complete question is.

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration through the board, and emerges with a speed of 345 m/sm/s.

What is the average acceleration of the bullet through the board/?

to know more about  the equation of motion :

brainly.com/question/13269040

#SPJ4

5 0
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Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m
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Answer:

18 N/C

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Using the relation for electric field due to dipole :

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E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

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I think it might be a Parasite
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