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lukranit [14]
3 years ago
13

Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a to

tal of 12 rubs, at a distance of 7.50 cm per rub, and with a frictional force of 45.0 N, what is the temperature increase in degrees Celsius?
Physics
1 answer:
Feliz [49]3 years ago
5 0

Answer:

Explanation:

Total heat = Work done = Force × distance

distance = 0.075 × 12 = 0.9 m

W = 45 × 0.9 = 40.5 joules

Specific heat of the human hand = 3.5 kj/kg = 3.5 j/g

Q = MCΔT

ΔT = (Q) ÷ (MC)

ΔT = 40.5 ÷ (3.5 × 1) = 11.57°C

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Answer:

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a).β=0.40 x10^{-4} T

Explanation:

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a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

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\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

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