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lord [1]
3 years ago
9

If the diameter of the space station is 730 m , how many revolutions per minute are needed for the "artificial gravity" accelera

tion to be 9.80m/s2?
Physics
1 answer:
nignag [31]3 years ago
4 0
For circular motion we know that

<span>F=ma=v^2/r </span>

<span>Therefore: </span>

<span>v = sqrt (rma) </span>

<span>Also, for cicular motion: </span>

<span>rev/min. = 60v/(2r*pi) </span>

<span>So your equation is: </span>

<span>rev./min = 60sqrt(rma)/(2pi*r) </span>

<span>For the mass (m) we can just use 1 kg. 
</span>
rev./min = 60sqrt(730*1*9.8)/(2pi*730) =60sqrt(7154)/(4584.4) 
rev./min = 60sqrt(7154)/(4584.4) =1.11 rev/min
<span>
the answer is </span>1.11 rev/min<span>

 </span>
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mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

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Pendulum 1 length = L₁ = 5 m

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Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

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f₁ = 0.22 Hz

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T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

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So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

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3 years ago
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Answer:

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