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Musya8 [376]
3 years ago
6

The speed of water at the tap of lower storey is more than that in the upper storey​

Physics
1 answer:
nadezda [96]3 years ago
4 0

Answer:

Pressure of liquid in container is given by, P= height × density × acceleration of gravity.

At the lower storey, the height of the liquid from the open end is great, since height is directly proportional to pressure, the pressure exerted by liquid is maximum hence increase in velocity of flow.

Unlike the upper storey where the height of water is less hence the pressure exerted by the liquid is minimum which decreases the velocity / speed of liquid flow

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Why would you expect the speed of light to be slightly less in the atmosphere then in a vacuum?
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The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material

<u>Explanation:</u>

When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.

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3 years ago
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3 years ago
If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
Mandarinka [93]

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No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

K.E=\frac{1}{2}mv^{2}

Similarly the momentum is given by m\times v

For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective kinetic energies is given by

K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}

K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}

Similarly For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective momenta are given by

p_{1}=m_{1}\times v_{1}

p_{2}=m_{2}\times v_{2}

Now since it is given that the two kinetic energies are equal thus we have

\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0
4 years ago
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