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rusak2 [61]
3 years ago
5

A camcorder has a power rating of 12 watts. If the output voltage from its battery is 3 volts, what current does it use?

Physics
2 answers:
serg [7]3 years ago
4 0

power=voltsxamps=watts.

12=3xamps

4 amps

------------

A. To increase the current, decrease the resistance.

----------------

r=v/i = 53/7

tatiyna3 years ago
4 0

Explanation:

The  expression for power in terms of voltage and current is as follows;

P=VI

Here, P is the power, V is the voltage and I is the current.

It is given in the problem that a camcorder has a power rating of 12 watts. The output voltage from its battery is 3 volts.

Calculate the current.

I= \frac{P}{V}

Put P= 12 W and V= 3 V.

I= \frac{12}{3}

I= 4 A

Therefore, the value of current is 4 A.

The expression for the Ohm's law is as follows;

V=IR                                                                                .......... (1)

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

It is given in the problem that  

The current in a hair dryer measures 13 amps. The resistance of the hair dryer is 9 ohms.

Calculate the voltage.

Put R= 9 ohms and I= 13 A.

V= (13)(9)

V= 117 V

Therefore, the value of the voltage is 117 V.

Ohm's law states that the potential difference is directly proportional to the current flowing across the conductor at given temperature. The current is inversely proportional to the resistance. To increase the current, decrease the resistance.

From the given option, the option (A) is correct.

It is given in the problem that a 1.5 m wire carries a 7 A current when a potential difference of 53 V is applied.

Calculate the resistance by using equation (1).

R= \frac{V}{I}

Put V= 53 V and I= 7 A.

R= \frac{53}{7}

R= 7.57 ohm

Therefore, the value of the resistance is 7.57 ohms.

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Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

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