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2 years ago

If a CD player uses a 12-V battery and draws 2.0 amps of current, how much power does it use?

Physics

1 answer:

IgorC [24]2 years ago

4 0

Answer:

The power is 24 watt .

Explanation:

Given that,

Voltage V= 12-V

Current I = 2.0 A

Using ohm's law

The current is directly proportional to the voltage.

In form of power,

The power is the product of the Current and voltage.

Formula of the power is defined as,

$P = I\times V$

$P = 2.0\times12$

$P = 24\ watt$

Hence, The power is 24 watt .

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ANOTHER PHYSICS QUESTION PLEASE HELP!

Sergio039 [100]

Answer

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3 years ago

Chose the word that correctly completes the following sentence: Kirchoff's junction law expresses the conservation of ____. a) m

Radda [10]

Answer:A- mass charge.

This can also be called current.

Explanation:

This is Kirchhoff’s 2nd law.

Kirchhoff’s junction law states that the sum of current(mass charge) flowing in and out of the junction must be equal to zero. This law emphasizes conservation of charge and energy. Charge is also a form of energy and it can neither be created nor destroyed.

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2 years ago

Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

$F_{x} + F{y} = ma$ (2)

The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

$F_{x} = 10 N - 4 N$

Notice that $f$ subtracts to $F$ since it is at the opposite direction.

$F_{x} = 6 N$

The forces present in y balance each other:

$F_{y} = 0$

Therefore:

$6 + 0 = ma$

$6 N = (2kg)a$ (3)

But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

So the block has an acceleration of $a = 3 m/s^{2}$ .

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3 years ago

You know what movie is rly good the great gastby and also ford v Ferrari

jolli1 [7]

Answer:HETJHETHRTHR

Explanation:

6 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

3 years ago