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Zolol [24]
3 years ago
15

A 1-kilogram parcel of air is at 35°c and contains 7 grams of water vapor. What is the relative humidity?

Physics
1 answer:
Nitella [24]3 years ago
8 0

Answer:

20%

Explanation:

Relative Humidity (%) = (water vapor content÷water vapor capacity) × 100

=(7÷35)×100

=(0.2)×100

=20%

According to the Temperature-Water Vapor Capacity Table, the water capacity at 35 °C is 35 grams.

Water Vapor Capacity: The amount of water (grams) which air can hold at a given temperature.

Water Vapor Content: The amount of water vapor actually present in the air.

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A 14-Ω coffee maker and a 14-Ω frying pan are connected in series across a 120-V source of voltage. A 20-Ω bread maker is also c
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Answer:

The total current supplied by the source of voltage = 10.29 A

Explanation:

We have a 14-Ω coffee maker and a 14-Ω frying pan are connected in series.

Effective resistance = 14 + 14 = 28Ω

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Effective resistance

             R=\frac{28\times 20}{28+20}=11.67\Omega

So we have resistor with 11.67Ω in a 120 V source of voltage.

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The total current supplied by the source of voltage = 10.29 A

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3 years ago
The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

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