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a famous relation in physics relates 39 moving m...

Question:
A famous relation in physics relates 'moving mass' m to the 'rest mass' m₀ of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

m = m₀ / (1-v²)^½

Answer:

Given the relation,

m = m₀ / (1-v²)^½

Dimension of m = M¹ L⁰ T⁰

Dimension of = M¹ L⁰ T⁰

Dimension of v = M⁰ L1 T^-1

Dimension of v² = M⁰ L2 T^-2

Dimension of c = M⁰ L1 T^-1

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1-v²)^½ is dimensionless i.e., (1 - v²) is dimensionless. This is only possible if v² is divided by c². Hence, the correct relation is

m = m₀ / (1-v²/c²)^½.

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Answered by Ekta Mehta
4 months ago

Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.

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Answered by Ekta Mehta
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Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.

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Answered by Ekta Mehta
4 months ago

Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.

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