Answer:
1626.4 N
Explanation:
Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?
The parameters to be considered are:
Distance S = 3m
Time t = 0.55s
Since the man started from rest, initial velocity u = 0
Using second equation of motion
S = Ut + 1/2at^2
3 = 1/2 × a × 0.55^2
3 = 1/2 × a × 0.3025
a = 3/ 0.15125
a = 19.83 m/s^2
Force = mass × acceleration
Force = 82 × 19.83
Force = 1626.4 N
Therefore, the force that water exerted on him is 1626.4 N
if currents go in opposite directions, wires repel
kinetic energy is usually measured in joule J which is equals to kgm²/s²
Answer:
D. 15 m/s downward
Explanation:
v = at + v₀
v = (-9.8 m/s²) (1.5 s) + (0 m/s)
v = -14.7 m/s
Rounded to two significant figures, the answer is D, 15 m/s downward.
