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Nikitich [7]
3 years ago
11

In Part H, you discovered that the luminosity of a light bulb increases if the current increases. The rate at which electric pot

ential energy is converted into heat depends on the current flowing through the bulb and the voltage across the bulb. This energy is supplied by the battery. Mathematically, the luminosity P of the light bulb is given by P=ΔVI, where ΔV is the voltage across the bulb and I is the current.What happens to the luminosity of the light bulb if the voltage of the battery is doubled? (Note that the PhET simulation does not display a numerical value for the luminosity, so you should use the relationship between the luminosity, the voltage across the bulb, and the current.)
Physics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

Bruh just do A bro thats the best answer...

Explanation:

Yes

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Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

P\cdot V = n \cdot R_{u}\cdot T (1)

Where:

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V - Volume.

n - Molar quantity, in moles.

R_{u} - Ideal gas constant.

T - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

P_{1}\cdot V_{1} = P_{2}\cdot V_{2} (2)

If we know that \frac{V_{2}}{V_{1}} = \frac{1}{3}, then the resulting pressure of the system is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 3\cdot P_{1}

The resulting pressure is 3 times the initial pressure.

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Answer:

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Explanation:

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