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eduard
3 years ago
5

If a cylindrical space station 275 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) mus

t it turn so that the outermost points have an acceleration equal to g ?
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

The centripetal acceleration is responsible for the artificial gravity because the acceleration of an object moving in constant circular motion causing from net external force is called centripetal acceleration. It defines to the center or seeking the center.

Given the following:

Cylindrical space station diameter           = 275 meters; 137.5 meters for the radius

Standard gravity                                       = 9.80665 m/s²

 

Using the formula:

w² x r =g

w² = g / r

w² = 9.80665 m/s² / 137.5 m

w² = 9.80665 m/s² / 137.5 m

w² = 0.0713 s²

Then take the roots

w = 0.267 this is radians per second / 2 x (3.1416 which is the pi)

w = 0.0424 rps convert to rpm

w = 0.0424 r/s (1minute / 60 seconds)

w = 7.08 x 10⁻⁴ revolutions per minute

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3 years ago
A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res
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The first thing you should know for this case is the definition of distance.
 d = v * t
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 v: speed
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8 0
3 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
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Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

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You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
3 years ago
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