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3 years ago

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(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 12

0 V (assume its resistance does not change), how bright will it be relative to 75-W 120-V bulbs? [Hint: Assume roughly that brightness is proportional to power consumed.]

1 answer:

Elodia [21]3 years ago

4 0

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

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Answer:

$\theta_2 - \theta_1 = 156.93 degree$

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

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$y = \frac{A}{5}$

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now we have

$\theta_1 = 11.53 degree$

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$\theta_2 - \theta_1 = 168.46 - 11.53$

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A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

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Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

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Q = -5.02× $10^{-9}$ C

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Area = A = (21.8× $10^{-2}$ m) (21.8× $10^{-2}$ m) = 4.75× $10^{-4}$ m²

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A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of

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Answer:

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Put the value into the formula

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We need to calculate the focal length

Using formula of magnification

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Put the value into the formula

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Using formula of for focal length

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