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ICE Princess25 [194]
3 years ago
10

If the voltage of the secondary side of the transformer is less than ? and the voltage of the primary side of the transformer ex

ceeds ? to ground, the secondary side shall be grounded.?
Physics
1 answer:
Natasha2012 [34]3 years ago
6 0

Answer:

50 V / 150 V

Explanation:

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Write a question about how changing temperature affects gas
Lera25 [3.4K]

Answer:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

Explanation:

One possible question can be:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:

V\propto T

Or also written as

\frac{V}{T}=const.

By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.

3 0
3 years ago
A beam of sunlight falling on a prism refracts and forms seven color bands. This illustrates that
ExtremeBDS [4]

That isn"t the right answer the correct answer is B.

4 0
3 years ago
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One of the light bulbs in this series circuit burns out, causing a break in the circuit.
ivanzaharov [21]

Answer:

It doesn't give light

Explanation:

No Flowing of electricity

3 0
2 years ago
The National Ambient Air Quality Standards (NAAQS) are maximum allowable levels for _____ harmful pollutants. sixteen six sixty
Harman [31]
6

I hope this helps (:

4 0
3 years ago
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What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?
Vlada [557]

Answer:

Ep=-2.3*10^{-19}J

Explanation:

The change in potential energy can be expressed as:

Ep=K.\frac{q1.q2}{r}

where K is a constant with a value of 9*10^{9}\frac{N.m^{2}}{C^{2}}, q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is +1.6*10^{-19}C and the charge for the electron is -1.6*10^{-19}C.

Converting r=1.0nm to m:

1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m

Replacing values:

Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}

Ep=-2.3*10^{-19}J

5 0
3 years ago
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