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katovenus [111]
1 year ago
15

A 10.0 Ω lightbulb is connected to a 12.0 V battery. (a) What current flows through the bulb? (b) What is the power of the bulb?

Physics
1 answer:
andreev551 [17]1 year ago
6 0

The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

<h3>What current flows through the bulb as well as the power of the bulb?</h3>

From ohm's law; V = I × R

Where V is the voltage, I is the current and R is the resistance.

Also, Power is expressed as; P = V × I

Where V is voltage and I is current.

Given that;

  • Resistance R = 10.0 ohms
  • Voltage V = 12.0V
  • Current I = ?
  • Power P = ?

First, we determine the current flow through the bulb.

V = I × R

12.0V = I × 10.0 ohms

I = 12.0 ÷ 10.0

I = 1.2A

Next, we determine the power of the bulb.

P = V × I

P = 12.0V × 1.2A

P = 14.4 Watts

Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

Learn more about Ohm's law here: brainly.com/question/12948166

#SPJ1

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Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

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s =  9.0 m

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Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

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Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

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a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w
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Answer:

40 cm

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We have to find the effort distance.

We know that

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800\times 10=200\times effort\;distance

Effort distance=\frac{800\times 10}{200}

Effort distance=\frac{8000}{200}

Effort distance=40 cm

Hence,  the effort distance will be 40 cm.

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