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mars1129 [50]
1 year ago
6

1 5kg cat is lifted 2 m into the air.how much gpe does it gain?

Physics
1 answer:
____ [38]1 year ago
6 0

Answer:

<h3><em><u>Potential Energy= mgh = (5)(9.8)(2)= 98.</u></em></h3><h3 />
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What are at least three types of energy involving a microwave
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8 0
3 years ago
yvonne van gennip of the netherlands ice skated 10.0 km with an average speed of 10.8 m/s. suppose vang ennip crosses the finish
Sophie [7]

By conservation of momentum, we will find that the mass is 4.97 kg.

So in the original system, we have two objects, the bouquet of flowers of mass M that is not moving and Yvonne, which has a mass of 63 kg and a speed of 10.8 m/s.

Then the total momentum of this system is:

P = (63kg)*(10.8m/s) + M*(0 m/s) = 680.4 kg*m/s.

Remember the conservation of momentum, thus, the final momentum must be equal to the above one.

In the final situation, Yvonne and the bouquet move together with a speed of 10.01 m/s,

Then the final momentum is:

P' = (63kg + M)*(10.01 m/s)

And that must be equal to the initial momentum, then we have the equation:

(63kg + M)*(10.01 m/s) = 680.4 kg*m/s.

630.63kg*m/s + M*10.01 m/s = 680.4 kg*m/s.

M*10.01 m/s = 680.4 kg*m/s - 630.63kg*m/s = 49.77 kg*m/s

M = (49.77 kg*m/s)/(10.01 m/s) = 4.97 kg

So the mass of the bouquet is 4.97 kg

If you want to learn more about momentum, you can read:

brainly.com/question/19636349

3 0
2 years ago
Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
Nana76 [90]

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

4 0
3 years ago
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