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Ilia_Sergeevich [38]
3 years ago
7

One particle has mass m and a second particle has mass 2m. The second particle is moving with speed v and the first with speed 2

v. How do their kinetic energies compare?
Physics
1 answer:
choli [55]3 years ago
3 0

Answer:

Explanation:

The formula for kinetic energy to be used here is 1/2mv².

If the first particle is "particle a" and the second particle is "particle n"; there kinetic energies (K.E) will be

K.Eₐ = 1/2.m2v² = mv²

K.Eₙ = 1/2.2mv² = mv²

From the above, <u>it can be said that there kinetic energies are the same</u>.

NOTE that the m and v used in the question means mass and velocity respectively.

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Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

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The phase at the second point is  \phi _2  = 1.57 \  rad

Explanation:

From the question we are told that

    The wavelength of the spherical wave is  \lambda =  2.0 \ mm =  \frac{2}{1000} = 0.002 \ m

    The first radius  is  r_1  = 4.0 \ mm  = \frac{4}{1000}  = 0.004 \ m

     The phase at that instant is  \phi _1 = \pi \ rad

     The second radius is  r_2  = 3.5 \ mm  = \frac{3.5}{1000}  = 0.0035 \ m

Generally the phase difference is mathematically represented as

          \Delta  \phi =  \phi _2 -  \phi _1

this can also be expressed as

         \Delta \phi =  \frac{2 \pi }{\lambda } (r_2 - r_1 )

So we have that

   \phi _2 -  \phi _1 =   \frac{2 \pi }{\lambda } (r_2 - r_1 )

substituting values

     \phi _2 -  \pi =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )

    \phi _2  =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) +   3.142

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