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3 years ago

We know that the motion of the Moon around the Earth is due

Physics

1 answer:

nadezda [96]3 years ago

3 0

Answer:

YES

Explanation:

Gravity acts as the centripetal force and the velocity earth has keeps it from falling on the sun.

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Determine the mechanical energy of this object: 1-kg ball rolls on the ground at 2 m/s

Fiesta28 [93]

The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...

KE =mv^2/2
KE= 1(2^2)/2 which the answer will come up by 2 Joules.

5 0

3 years ago

What is the definition of power ? what are the units of power?

Jet001 [13]

Answer:

power is the rate at which energy is transferred or converted

the unit of power is Watts

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3 years ago

What is the average speed of a race car that moved 20 kilometers in 10 minutes?

Nadya [2.5K]

Answer:21

Explanation:every body said

8 0

3 years ago

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

$v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)$

The final speed of the leading car is given by the following formula:

$v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)$

4 0

3 years ago

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

6 0

3 years ago