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Murljashka [212]
3 years ago
14

A 65.0-kg woman steps off a 11.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 4.60 m,

what is the average resistance force exerted on her by the water
Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

2160.26 N.

Explanation:

From the question,

The total energy of the woman = work done by the water.

mg(h+d) = F'×d.......................... Equation 1

m = mass of the the woman, g= acceleration due to gravity, h = height of the platform, d = depth of water reached by the woman, F'= Resistance force exerted on her by the water.

make F' the subject of the equation

F' = mg(h+d)/d............................ Equation 2

Given: m = 65.0 kg, g = 9.8 m/s², h = 11 m, d = 4.6 m.

Substitute into equation 2

F' = 65(9.8)(11+4.6)/4.6

F' = 9937.2/4.6

F' = 2160.26 N.

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