Answer:
31.905 ft/s²
Explanation:
Given that
Mass of the pilot, m = 120 lb
Weight of the pilot, w = 119 lbf
Acceleration due to gravity, g = 32.05 ft/s²
Local acceleration of gravity of found by using the relation
Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)
119 = 120 * a/32. 174
119 * 32.174 = 120a
a = 3828.706 / 120
a = 31.905 ft/s²
Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²
I would say option D, it depends on the size of the star
<span>Make the surfaces smoother. Rough surfaces produce more friction and smooth surfaces reduce friction
Lubrication is another way to make a surface smoother
Make the object more streamlined
Reduce the forces acting on the surfaces
<span>Reduce the contact between the surfaces.</span></span>
The 1st one is basically B because it will stay in motion with the same speed and in the same direction unless acted on by an unbalanced force and the 2cd one is A because most of is transformed into thermos energy. hope this helps!
