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USPshnik [31]
3 years ago
8

The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular mo

tion with the radius of your circular motion being equal to the radius of the Earth. You are an evil mad scientist and have come up with the simultaneously awesome and terrifying plan to increase the speed of the Earth’s rotation until people at the Earth’s equator experience a centripetal (radial) acceleration with a magnitude equal to g, (9.81 m/s2 ), effectively making them experience weightlessness. If you succeed in your dastardly plan, what would be the new period of the Earth’s rotation?
a. 2.7 minutes b. 84 minutes c. 48 minutes d. 76 minutes
Physics
1 answer:
emmainna [20.7K]3 years ago
7 0

Answer:

b. 84 minutes

Explanation:

a_c=g = Centripetal acceleration = 9.81 m/s²

r = Radius of Earth = 6378.1 km

v = Velocity

Centripetal acceleration is given by

a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s

Time period is given by

T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes

The time period of Earth’s rotation would be 84.43835 minutes

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vovangra [49]

Answer:

The speed of the particle is 34.66 m/s.

Explanation:

Given that,

Mass of particle = 50 g

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Time = 5.0 sec

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v=a\times t

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v=\dfrac{4.0\times10^{-3}\times60\times5.0}{50\times10^{-3}}

v=24\ m/s

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v=\sqrt{v_{x}^2+v_{y}^2}

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v=\sqrt{25^2+24^2}

v=34.66\ m/s

Hence, The speed of the particle is 34.66 m/s.

6 0
3 years ago
One billiard ball is shot east at 1.8 m/s . A second, identical billiard ball is shot west at 0.80 m/s . The balls have a glanci
ycow [4]

Answer:

Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.

Explanation:

By conservation of the linear momentum:

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What is the connection between the x- and y-motions of a projectile?
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Answer c, velocity would be the answer.
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4 0
3 years ago
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