Question #e2ec5

Question

1136 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-27T05:41:40

.

P = force/area #=(kgms^-2)/m^2= kgm^-1s^-2#
v#= ms^-1#
A#=m^2#

#Q= (kgm^-1s^-2)/(ms^-1 m^2)#

#Q=kgm^(-1-3)s^(-2-(-1)#

#Q=kgm^-4s^-1#

Answer 2 · 2017-05-27T05:42:12

#Kgm^-4sec^-1#

See , ATQ

#Q=(P)/(v*A)#
#=>(F/A)/(m/s*A)#
#=>(F)/(v*A^2)#
#=>(MLT^(-2))/(LT^(-1)*L^5)#

#=>(MLT^-2)/(L^5T^-1)=ML^-4T^-1#

hence , equals to #Kgm^-4sec^-1#