Aperson pushes a 16.0-kg shopping cart at a constant velocity for a distance of 16.0 m. she pushes in a direction 30.0° below the horizontal. a 54.0-n frictional force opposes the motion of the cart. (a) what is the magnitude of the force that the shopper exerts? n (b) determine the work done by the pushing force. j (c) determine the work done by the frictional force. j (d) determine the work done by the gravitational force. j

Let: 
P - pushing force, 
t - angle below horizontal, 
F -  friction force. 

Resolving horizontally: 
P cos(t) = F 

(a) 
P = F / cos(t) 
= 54  / cos(30.0) 
= 62.35 

(b) 
Work done by P 
= P cos(t) * 16.0 
= F * 16.0 
= 54.0 * 16.0 
= 864J 

(c) 
Work done by F 
= - F * 16.0 
= - 54 * 16.0 
= - 864 J. 

(d) 
As there is no motion vertically, the work done by the gravitational force is 0.

your answer is, orbit around the nucleus

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr