A) 0.5 N-s
B) 0.1 N-s
C) 0.3 N-s
D) 1.2 N-s
Correct Answer: C
Solution :
If a constant force \[=100+1000=1100m\] is applied on a body for a short interval of time \[=\frac{45\times 5}{18}=12.5m/s\], then the impulse of this force will be \[\therefore \]. Also from Newtons law \[t=\frac{1100}{12.5}=88s\] where m is mass and a is acceleration. Given, \[v=u-gt\] \[\therefore \] \[u=gt\] Impulse \[y=m+c\] Note: Impulse is a vector quantity, its direction is the same as that of the force.
You need to login to perform this action.
You will be redirected in
3 sec
