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GalinKa [24]
3 years ago
15

What would happen if the Earth stopped rotating?

Physics
1 answer:
trapecia [35]3 years ago
4 0

Answer:

the atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. ... This means rocks, topsoil, trees, buildings, your pet dog, and so on, would be swept away into the atmosphere.

Explanation:

to me, this means we would proabaly be sucked into outer space and could die if no astronaut gear is on

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What is the most likely elevation of point B?<br>a. 150 ft<br>b. 200 ft<br>c. 125 ft<br>d. 225 ft​
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Answer:

D

Explanation:

erm i think its 225 ft?

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According to Newton's Third Law, what is the equal and opposite force to the downward force of gravity pulling on a man standing
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The equal and opposite reaction to the gravitational force of the Earth pulling down on a man standing on the beach is the gravitational force of the man pulling up on the Earth. <em>(A)</em>

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A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the c
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Answer:

<em>The velocity of the truck is 3.33 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If some collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

There are two objects: The m1=4000 Kg car and the m2=6000 Kg truck. The car was moving initially at v1=4 m/s and the truck was at rest v2=0. After the collision, the car moves at v1'=-1 m/s. We need to find the velocity of the truck v2'. Solving for v2':

\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}

Substituting:

\displaystyle v'_2=\frac{4000*4+6000*0-4000(-1)}{6000}

\displaystyle v'_2=\frac{16000+4000}{6000}

\displaystyle v'_2=3.33

The velocity of the truck is 3.33 m/s

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