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sveta [45]
3 years ago
6

A collar A of mass m is moving with a speed v when it strikes an identical collar B that is at rest. Knowing that the coefficien

t of restitution between the collars is e. Identify the correct relation for the energy lost in the impact as a function of m, e, and v.

Physics
1 answer:
klio [65]3 years ago
8 0

Answer:

ΔK.E = mv²(1-e²)

Explanation:

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In which situation would an object weigh the LEAST? (assume all the objects have the same mass)
inessss [21]

Answer:

An object on the moon would weigh the LEAST among these. So correct answer is B.

Explanation:

  • Weight of an object on any place is given by:

W = Mass * Acceleration due to gravity(g)

  • It means when masses of different objects those are in different places are same, the weight of  those objects depends upon the 'g' of that particular place.
  • As we know, acceleration due to gravity on surface of moon (g') is 6 times weaker than the acceleration on surface of earth (g), which is due to the large M/R^2 of the earth than the moon.

i.e. g' = g/6 so W' = W/6

  • And in the space between the two, the object is weightless.
8 0
3 years ago
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marshall27 [118]

Answer:

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4 0
2 years ago
As the wave interacts with a wall, which kind of wave interaction is shown?
kow [346]
 The answer would be a reflection. This is because, t<span>he color of an object is actually the wavelengths of the light reflected while all other wavelengths are absorbed. Color, in this case, refers to the different wavelengths of light in the </span>visible light spectrum<span>perceived by our eyes. The physical and chemical composition of matter determines which wavelength (or color) is reflected.</span>
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3 years ago
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Hi :) how to do this?
fredd [130]

Answer:

The answer is b

Hope this helps you!

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Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.
Mars2501 [29]

Answer:

Rms speed of the particle will be  38.68\times 10^8m/sec

Explanation:

We have given mass of the air particle m=5\times 10^{-16}kg

Gas constant R = 8.314 J/mol-K

Temperature is given T = 27^{\circ}C=273+27=300K

We have to find the root mean square speed of the particle

Which is given by v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec

So rms speed of the particle will be 38.68\times 10^8m/sec

5 0
3 years ago
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