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Nikitich [7]
4 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]4 years ago
4 0

Answer:

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

Explanation:

Yes

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ivanzaharov [21]
<span>anwser will be 

F = ma

where 

F = force exerted on the bullet 
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Vf^2 - Vo^2 = 2as 

where 

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given) 
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Substituting appropriately, 

326^2 - 0 = 2(a)(0.83) 

a = 64,022 m/sec^2 

the anwser will be
Substituting this into Equation 1, 

F = 0.005(64,022) 

F =320.11 Newtons 

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