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belka [17]
3 years ago
5

5) A 80-kg man has a total foot imprint area of 480 cm2. Determine the pressure this man exerts on the ground if (a) he stands o

n both feet and (b) he stands on one foot.
Engineering
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

The pressure exerted by this man on ground

(a) if he stands on both feet is 8.17 KPa

(b) if he stands on one foot is 16.33 KPa

Explanation:

(a)

When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:

Area = A = 2 x 480 cm²

A = 960 cm²

A = 0.096 m²

The force exerted by man on his area will be equal to his weight.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.096 m²

<u>P = 8166.67 Pa = 8.17 KPa</u>

(b)

When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:

Area = A = 480 cm²

A = 0.048 m²

The force exerted by man on his area will be equal to his weight, in this case, as well.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.048 m²

<u>P = 16333.33 Pa = 16.33 KPa</u>

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we know that, the relationship between turn and impedance is

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400 / 25 = 900² / N₂²

we cross multiple to get our N₂

400 × N₂² =  900² × 25

N₂² = ( 900² × 25 ) / 400

N₂² = ( 810000 × 25 ) / 400

N₂² = 20250000 / 400

N₂² = 50625

N₂ = √50625

N₂ = 225

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