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Ilia_Sergeevich [38]
4 years ago
12

Light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb

Physics
1 answer:
masya89 [10]4 years ago
5 0

Answer:

<h2>121ohms</h2>

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

<em>Hence, the resistance of this bulb is 121 ohms</em>

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To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

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Let the third charge be q_3 = +Q is placed at a distance x from q_1

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F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}

The condition of equilibrium is

F_{1,3} = F_{2,3}

\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}

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x = 0.331 from q_1

To find the magnitude of q_3 we use F_{1,2} = F_{1,3}

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uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work i
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Answer:

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work done in stretching the spring from 30 cm to 45 cm, W = 3 J

extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m

The work done is given by;

W = ¹/₂kx²

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k is the force constant of the spring

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W = ¹/₂kx²

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