Question

if Momentum P and velocity V are related as P is equals to under root2 divided by b + bsquare then dimensional formula of e to the power minus 2 is​

Answer 1

Answer:

Dimension of the given physical quantity is given as

$ab^{-2} = M^2L^2T^{-2}$

Explanation:

As we know that

$P = \frac{\sqrta}{b + v^2}$

Now we know that two physical quantities of same dimension can be added together

so we have

dimension of b = dimension of v^2

so we have

$b = M^0L^2T^{-2}$

now we have this whole equivalent to the dimension of momentum

so we have

$\frac{\sqrta}{M^0L^2T^{-2}} = M^1L^1T^{-1}$

so we have

$a = M^2L^6T^{-6}$

now we have to find

$ab^{-2}$

$(M^2L^6T^{-6})(L^{-4}T^4)$

so we have

$ab^{-2} = M^2L^2T^{-2}$

#Learn

Topic : Dimension

https://brainly.in/question/3631086