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3 years ago

What is the current in the 10.0 , resistor?

Physics

2 answers:

In-s [12.5K]3 years ago

5 0

Answer:

B. 12.0A

Explanation:

The circuit shown in the image is a parallel circuit which mean that the voltage of the source is equal across each resistance in the circuit because they are directly connected to the terminals of the voltage source.

The voltage across the 10 Ohms resistor is 120V

So, let's proceed to calculate the current

By using Ohm's Law

$I=\frac{V}{R} \\I= \frac{120.0V}{10.0Ohm} \\I= 12.0A$

12A is the amount of current flowing through the 10 ohms resistor.

olga nikolaevna [1]3 years ago

4 0

Current = (voltage) / (resistance)

The ends of the 10-ohm resistor are connected directly to both terminals of the battery.

So the voltage across the resistor is the voltage of the battery ... 120 V. None of that other stuff in the circuit has any effect on it.

Current = (120 V) / (10 ohms)

<em>Current = 12 A . (B)</em>

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Explanation:

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A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi

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Explanation:

Given

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radius of wheel $r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm$

mass of clay $m_2=2.8\ kg$

the radius of the chunk of clay $r_2=8\ cm$

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Combined moment of inertia of wheel and clay chunk

$I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2$

Conserving angular momentum

$\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi$

Common frequency of wheel and chunk of clay is

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Answer:

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