Answer:
2,500 feet (760 meters)
Explannation: <em>At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.</em>
Answer/solution:
Given :
Mass =5kg
T 1 =20 C,T 2 =100 ∘C
ΔT=100−20=80 ∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
The answer is D.
I hope this helps.
