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True [87]
3 years ago
12

A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V house

hold electricity, decreasing the resistance of the filament will _____ the current through the bulb and _____ the power dissipated by the bulb. A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V household electricity, decreasing the resistance of the filament will _____ the current through the bulb and _____ the power dissipated by the bulb. Decrease, decrease Decrease, increase Increase, increase Increase, decrease
Physics
1 answer:
Igoryamba3 years ago
8 0

Answer:

Increases, increases

Explanation:

The current is directly proportional to the voltage and inversely proportional to the resistance. The implication of this is that, whenever the voltage is increased, the current increases simultaneously. On the other hand, if the resistance is increased, the current will decrease accordingly and vice versa.

Recall that power is given by P= V^2/R where;

P= power, V= voltage and R= resistance

We can see that power and resistance are inversely related hence decreasing the resistance increases the power output of the lightbulb.

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What happen to the frequency of transverse vibration of a stretched string if its tension is halved and the area of cross sectio
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Answer:

The fundamental frequency of the stretched string is:

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Here,

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If T is halved and A is doubled,

f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f

Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.

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Answer:

28.796 J

Explanation:

I hope that it's a correct answer.

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