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 the light c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy records the relation almost correctly but forgets way to put the constant c. He writes:
Guess where to put the missing 'c'.
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Hii friend,
# Step-by-step solution-
Given incomplete equation is,
m = m0 / (1-v^2)^0.5
(1-v^2)^0.5 = m0/m
On RHS , [M]/[M] = null
By principle of homogenicity of dimensions every term in eqn. must have same dimensions.
As RHS is dimensionless, LHS must be so.
(1-v^2)^0.5 = [null]
To make it so v have to be divided by c.
Therefore, formula should have been like
m = m0 / [1-(v/c)^2]^0.5
Hope This helped someone...
# Step-by-step solution-
Given incomplete equation is,
m = m0 / (1-v^2)^0.5
(1-v^2)^0.5 = m0/m
On RHS , [M]/[M] = null
By principle of homogenicity of dimensions every term in eqn. must have same dimensions.
As RHS is dimensionless, LHS must be so.
(1-v^2)^0.5 = [null]
To make it so v have to be divided by c.
Therefore, formula should have been like
m = m0 / [1-(v/c)^2]^0.5
Hope This helped someone...
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