3 For this force system the equivalent system at P is A FRP 40 lb along x dir and MRP 60 ft lbB FRP 0 lb and MRP 30 ft lbC FRP 30 lb along y dir and MRP 30 ft lbS FRP 40 lb along x dir and MRP 30 ft lb

This question is incomplete, the complete question is;

For this force system the equivalent system at P is ___________

A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft.lb

B) FRP = 0 lb and MRP = +30 ft.lb

C) FRP 30 lb (along +y-dir.) and MRP  = -30 ft.lb

D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb

D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb

Explanation:

From the figure in the image i uploaded along this answer;

FRP = ( 40 lb i + 30 lb j ) + [30 lb (-j)]

Where i  and j are the unit vectors along X & Y axis respectively.

So, FRP = 40 lb i

that is,  FRP = 40 lb along +X direction

MRP = [ 30 lb x ( 1 ' + 1' ) ] +( -30 lb x 1 ' )

= (30 lb x 2 ' )- 30 lb ft

= 60 lb ft - 30 lb ft

= 30 lb ft

Therefore option(D) is correct    

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j