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NeX [460]
2 years ago
13

A car with a mass of 500 kg is moving at a speed of 12 m/s. How much kinetic energy does it have?

Physics
1 answer:
shtirl [24]2 years ago
5 0

Answer:

36000

Explanation:

1/2mv²=0.5×500×12²=36000j

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What happens to the convection fluid as it heats up? How might this relate to masses of heated air?
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Answer:

The convection process plays an important role in the liquid. Due to the increasing heat supply or high amount of temperature, the fluid gets heated up, as a result of which it becomes warm, less dense and eventually rises up forming convection cells.

In the interior of the earth, the hot molten rocks get heated up due to the heat supplied by the core of the earth. This makes the magma warm and less dense and rises upward forming convection currents in the mantle.

This convection process is similar to the convection cells that form in the atmosphere, where the hot, less dense air rises up in the atmosphere forming a low-pressure zone. This uprising air forms convection cells, in which the warm air rises and as it rises high in the atmosphere, the temperature becomes low, making the air cold and it eventually sinks.

8 0
3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
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Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
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What is the length of a simple pendulum with a period of 2.0 s?
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T=2.pie sqr root(l/g)

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