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snow_lady [41]
3 years ago
14

2. An 85.5 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 12

.0 m) and spun at a constant angular velocity of 18.4 rpm. Answer the following:
(a) What is the angular velocity of the centrifuge in ?
(b) What is the linear velocity of the astronaut at the outer edge of the centrifuge?
(c) What is the centripetal acceleration of the astronaut at the end of the centrifuge?
(d) How many g’s does the astronaut experience?
(e) What is the centripetal force experienced by the astronaut?
Physics
1 answer:
icang [17]3 years ago
8 0

Answer Expert Verified

Hello here is your answer and also please mark me as brainlest

Since one revolution around a circle is 2 pi radian; hence 1 rpm equals 2 pi radians per minute. And because a minute has 60 seconds, 1 rpm equals 1/60 revolution per second. Therefore, we have 2 pie/60 * 15.3 = 0.2513 rps. The linear velocity v = wr where w is the angular velocity in rad/s and r is the distance. So we have 0.2513 * 10.0 = 2.513 rad/s The centripetal acceleration is given by a = w^2 r = (2.513)^2 * 10 = 63.15 rad/s2 The centripetal force F = mass * centripetal acc = 75 * 63.15 = 4736.25 N The torque = centripetal force * distance = 4736 * 63.15 = 299078.4 Nm Two forces acts on the astronaut. The normal force and acceleration due to gravity.

I hope this help you

brainlest

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Explanation:

The volume does not affect the potential energy, as this energy depends on the mass and elevation of the body relative to the reference point. This analysis can be easily seen in the equation expressing potential energy

E_{p} =m*g*h\\where:\\m=mass[kg]\\g=gravity[m/^2]\\h=elevation[m]

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As per Bernuolli's Theorem total energy per unit mass is given as

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Part B)

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7 0
4 years ago
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Blababa [14]

Answer:

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