PHYSICS HIGH SCHOOL

A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Answers

Answer 1

Answer:

The angular velocity of the disk must be 2.25 rpm

Explanation:

The centripetal acceleration of an object in circular motion is given by

$a=\omega^2 r$

where

$\omega$ is the angular velocity

r is the distance of the object from the axis of rotation

For the space station in this problem, we have

$a=(g)/(2)=(9.8)/(2)=4.9 m/s^2$ is the centripetal acceleration

The diameter of the disk is

d = 175 m

So the radius is

$r=(175)/(2)=87.5 m$

So, a point on the rim has a distance of 87.5 m from the axis of rotation. Therefore, we can re-arrange the previous equation to find the angular velocity:

$\omega = \sqrt{(a)/(r)}=\sqrt{(4.9)/(87.5)}=0.237 rad/s$

And this is the angular velocity of any point along the disk. Converting into rpm,

$\omega=0.236 (rad)/(s)\cdot (60 s/min)/(2\pi rad/rev)=2.25 rpm$

Learn more about circular motion:

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Answers

The answer to the given question is lightning.
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A gyre is a set of currents that form:a. a line
b. a loop
c. weather
d. climate zones

Answers

A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.

Answer:

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An Olympic sprinter can go from a state of rest to 11 meters per second in 10 seconds. What is the average acceleration of the sprinter? A.
0.9 m/s

B.
110 m/s

C.
110 m/s²

D.
1.1 m/s²

Answers

Answer

Average accleration is 1.1 m/s² .

Option (D) is correct .

Explanation:

Formula

$Average\ accleration = (v_(final)-v_(initial))/(t_(final)-t_(initial))$

As given

An Olympic sprinter can go from a state of rest to 11 meters per second in 10 seconds.

$v_(initial) = 0\ meter\ per\ second$

$v_(final) = 11\ meter\ per\ second$

$t_(initial) = 0\ second$

$t_(final) = 10\ second$

Putting all the values in the formula

$Average\ accleration = (11-0)/(10-0)$

$Average\ accleration = (11)/(10)$

Average accleration = 1.1 meter per second²

Therefore Average accleration is 1.1 m/s² .

Option (D) is correct .

an Olympic sprinter can go from a state of rest to 11 meters per second in 10 seconds. 1.1 m/s² is the average acceleration of the sprinter

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D) Good conductors

Answers

Answer: D
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Answers

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Answers

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