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aleksandr82 [10.1K]
3 years ago
9

Championship swimmers take about 22 s and about 30 arm strokes to move through the water in a 50 m freestyle race.The swimmer's

metabolic power is 800 W . A) If the efficiency for swimming is 25 %, how much energy is expended moving through the water in a 50 m race? B) If half the energy is used in arm motion and half in leg motion, what is the energy expenditure per arm stroke? C)Model the swimmer's hand as a paddle. During one arm stroke, the paddle moves halfway around a 90 cm-radius circle. If all the swimmer's forward propulsion during an arm stroke comes from the hand pushing on the water and none from the arm (somewhat of an oversimplification), what is the average force of the hand on the water?
Physics
1 answer:
Tasya [4]3 years ago
3 0

Answer:

4400 Joules

73.33 Joules

25.9325 Joules

Explanation:

P = Power = 800 W

t = Time = 22 s

F = Force

r = Radius of arm = 90 cm

Energy

E=P\times t\\\Rightarrow E=800\times 22\\\Rightarrow E=17600\ J

As the efficiency is 25%

E=0.25\times 17600\\\Rightarrow E=4400\ J

Energy used in the race is 4400 Joules

Half of the energy is used in the arm

E=\frac{4400}{2}=2200\ J

So, per stroke of paddle

E=\frac{2200}{30}=73.33\ J

The energy expenditure per arm stroke is 73.33 Joules

Displacement will be the half of the perimeter of the circle

s=\frac{1}{2}\times 2\pi r\\\Rightarrow s=\frac{1}{2}\times 2\pi 0.9\\\Rightarrow s=2.82743

Work done

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.33}{2.82743}\\\Rightarrow W=25.9352\ J

The average force of the hand on the water is 25.9325 Joules

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Answer:

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Explanation:

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Two loops of wire are arranged so that a changing current in one will induce a current in the other. If the current in the first
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Answer:

The current in the second loop will stay constant

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WILL MARK BRAINLIEST
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Answer:

aww that's simple! its B can i get brainliest? i really need some points rn <3

Explanation:

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A cliff diver from the top of a 100 (m) cliff. He begins his dive by jumping up with a velocity of 5 (m/s) a. How
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Answer:

A.) 4.81 seconds

B.) 44.6 m/s

Explanation:

He begins his dive by jumping up with a velocity of 5 (m/s).

Let us first calculate the maximum height reached by using third equation of motion

V^2 = U^2 - 2gH

At maximum height, V = 0

0 = 5^5 - 2 × 9.8H

19.6H = 25

H = 25 /19.6

H = 1.28 m

The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.

Where height h = 1.28 + 100 = 101.28 m

h = Ut + 1/2gt^2

As the diver drop from maximum height, U = 0

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4.9t^2 = 101.28

t^2 = 101.28/4.9

t^2 = 20.669

t = sqrt ( 20.669)

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The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds

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