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3 years ago

In your own words, what is the relationship between temperature, dew point temperature and air pressure

2 answers:

8 0

The dew point is the temperature the air would be at if it saturated at the present level of water vapour content. So dew point is related to air volume which is related to both temperature and pressure. Temperature would be the heat of the air around you. And air pressure <span>is the force exerted onto a surface by the weight of the </span>air<span>. I hope I could help clarify it a little bit for you :D

</span>

WITCHER [35]3 years ago

7 0

<span>The dew point is the temperature below which the water vapor in a volume of humid air at a constant barometric pressure will condense into liquid water.

Condensed water is called dew when it forms on a solid surface.

Brainliest if satisfied :D
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A point source of light is located at the bottom of a steel tank, and an opaque circular card of radius is placed horizontally o

pochemuha

Start by using the addition as a sign and use multiplying

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3 years ago

The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it

horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

Where P = Power, I = rms current, V = rms voltage.

make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

V = 1500(4.525)

V = 6787.5 V

Hence the rms voltage = 6787.5 V

7 0

3 years ago

Which factors describe a gas

Alekssandra [29.7K]

Answer:

B. changing shape and changing volume

Explanation:

*no definite shape (takes the shape of its container)

*no definite volume

7 0

4 years ago

If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is

saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 0

3 years ago

Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The sec

SpyIntel [72]

Answer:

2.32
0.43

Explanation:

12.0 kv primary voltage

315 kv secondary voltage ( converted voltage ) V1 or Vo

v2 (Vn)= 730 kv new secondary voltage

a) Ratio of turns in 730 kv to turns in 315 kv

$\frac{Vn}{Vo} = \frac{Nn}{No}$ = $\frac{730}{315}$ therefore the ratio of turns = 2.317 ≈ 2.32

B) ratio of the new current output to the old current output for the same power input to the transformer

since the power input is the same

$\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }$ equation 1

Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current

therefore equation 1 becomes

$\frac{In}{Io} = \frac{Vo}{Vn}$ = 315 / 730 = 0.43

7 0

3 years ago