Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
Weight increases but mass stays the same

Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then

Solving for
we get

Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is

The exhaust rate dm/dt is


Therefore, the velocity at which the exhaust gases exit the engines is

