Answer:
The terminal velocity of the diver is 115 m/s = 414 km/hr
Explanation:
At terminal velocity,
Fnet = mg - Fd = 0
Drag force, Fd = cρAv²/2
mg = cρAv²/2
Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by
v = √(2mg/ρcA)
where m = mass of body falling through fluid = 80 kg
g = acceleration due to gravity = 9.8 m/s²
ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³
c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7
A = the area of the diver facing the fluid = 0.14 m²
v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr
Given that,
Mass of the stone, m = 400 g = 0.4 kg
Initial speed, u = 20 m/s
It is climbed to a height of 12 m.
To find,
The work done by the resistance force.
Solution,
Let v is the final speed. It can be calculated by using the conservation of energy.

Work done is equal to the change in kinetic energy. It can be given as follows :

So, the required work done is 32.99 J.
Answer:
Explanation:
Ask a question that can be answered by making observations.
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