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Alex777 [14]
3 years ago
8

A man consumes 2055 kcal of food in one day, converting most of it to body temperature. If he loses half this energy by evaporat

ing water (through breathing and sweating), how many kilograms of water evaporate?
Physics
2 answers:
vitfil [10]3 years ago
8 0

Answer:

1.903 kg.

Explanation:

Energy obtained from the food, Q= 2055kCal

Half of the energy lost to evaporating water = 2055/2

= 1027.5 kCal

Latent heat of water, L = 540 kCal/kg

Q= m × L

m = 1027.5/540

= 1.903 kg.

9966 [12]3 years ago
3 0

Answer:

3.81kg

Explanation:

Given that heat lost by man Q= 2055kCal

Latent heat of water L = 540KCal/Kg

Mass of water evaporated = M

Using

Q= ML

2055= M * 540

M= 2055/540

M= 3.81kg

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3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
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Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

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  • r is the distance
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Let electric field due to charge q be F1 and -q be F2

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⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

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