Question

calculate the dimensions of force and impulse taking velocity, density and frequency as basic quantities.

Answer 1

Answer:

The dimension of force and impulse is $F= k v^{4}\rho^{1}f^{-2}$ and $I= k v^{4}\rho^{1}f^{-3}$.

Explanation:

We know that,

Let the force is directly proportional to velocity, frequency and density.

$F\propto v^{x}$

$F\propto f^{y}$

$F\propto \rho^{y}$

$F=k v^{x}\rho^{y}f^{z}$

The dimension formula of force,velocity, frequency and density.

$F=[MLT^{-2}]$

$v=[LT^{-1}]$

$\rho=[ML^{-3}]$

$f=[T^{-1}]$

$[MLT^{2}]= [LT^{-1}]^{x}[ML^{-3}]^{y}[T^{-1}]^{z}$

$[MLT^{2}]=[M^{y}L^{x-3y}T^{-x-z}]$

$M^{1}=M^{y}$

$y= 1$

$L^{1}=L^{x-3y}$

$x-3y=1$

$x= 4$

$T^{-2}=T^{-x-z}$

$z=-2$

Now, The dimension of force will be

$F= kv^{4}\rho^{1}f^{-2}$

The dimension formula of impulse

$I=[MLT^{-1}]$

$[MLT^{-1}]= [LT^{-1}]^{x}[ML^{-3}]^{y}[T^{-1}]^{z}$

$[MLT^{-1}]=[M^{y}L^{x-3y}T^{-x-z}]$

$M^{1}=M^{y}$

$y= 1$

$L^{1}=L^{x-3y}$

$x-3y=1$

$x= 4$

$T^{-1}=T^{-x-z}$

$z=-3$

Now, The dimension of impulse will be

$F= kv^{4}\rho^{1}f^{-3}$

Hence, The dimension of force and impulse is $F= k v^{4}\rho^{1}f^{-2}$ and $I= k v^{4}\rho^{1}f^{-3}$.

Answer 2

Explanation:

kv4ρ1f−2 and I= k v^{4}\rho^{1}f^{-3}I=kv4ρ1f−3 .