Answer:
3.416 m/s
Explanation:
Given that:
mass of cannonball 
= 72.0 kg
mass of performer 
= 65.0 kg
The horizontal component of the ball initially 
= 6.50 m/s
the final velocity of the combined system v = ????
By applying the linear momentum of conservation:
v = 3.416 m/s
The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop.
Hope this helps!
During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s
Best Answer:<span> </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km
The work done by the aircraft overcoming the wind is
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>
