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3 years ago

The useful energy from a computer is carried away from the computer by sound and what else?

Physics

1 answer:

Effectus [21]3 years ago

4 0

light

Explanation:

a computer has a screen and speakers, which produce light and siund

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What does the number, or density, of field lines on a source charge indicate?

Harrizon [31]

Hello.

The answer: C. amount of charge on the source charge

The field lines all represent the movements of field charges. The more crowded the field lines or the more density of electric field the higher is the magnitude of the source charge.

Have a nice day.

6 0

3 years ago

Read 2 more answers

Which property of sound waves decreases as the square of the distance from the source increases?

Vitek1552 [10]

Answer:

Intensity

Explanation:

The intensity of a sound wave is equal to the ratio between to the power emitted by the source divided by the area of the spherical surface through which the wave propagates:

$I=\frac{P}{4\pi r^2}$

where

P is the power

$4\pi r^2$ is the area of the spherical surface

r is the distance from the source

As we see from the formula, the intensity is inversely proportional to the square of the distance from the source:

$I\propto \frac{1}{r^2}$

so, intensity is the correct answer.

3 0

3 years ago

Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge

butalik [34]

Answer:

Electric charge in the earth will be $Q=5.231\times 10^5C$

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}$ . here K is coulomb's constant

So $116=\frac{9\times 10^9\times Q}{(6371000)^2}$

$Q=5.231\times 10^5C$

So electric charge in the earth will be $Q=5.231\times 10^5C$

5 0

3 years ago

What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?

AnnZ [28]

V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
F = 0.08810409956 Hz

4 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago