Question

Problem=
The acceleration (a) of a particle is given by
a =
$\alpha fx + \beta {}^{2} t \frac{}{ \alpha + n}$
find the Dimension of
$\frac{ \beta {}^{2} }{ \alpha + n}$


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Answer 1

Answer:

The acceleration of the particle as a function of x , is given by. ... to v(x)=βx−2n, where β and n are constants and x is the position of the particle. ... a(x)=βx−2nx−2n×β(−2n)x−2n− 1.