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2 years ago

9

How are the amplitude and energy of a wave related

2 answers:

vovangra [49]2 years ago

4 0

B is most likely the answer (Wave amplitude and energy are not related)

In-s [12.5K]2 years ago

3 0

I would go with the last two because they are the same please give me brainiest if i am right

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40 POINTS!!! + BRAINLIEST!!!

Sergeu [11.5K]

Answer:

I think it is better if you read and shortly write my explanation

Explanation:

simple pendulum with no friction, mechanical energy is conserved. Total mechanical energy is a combination of kinetic energy and gravitational potential energy. As the pendulum swings back and forth, there is a constant exchange between kinetic energy and gravitational potential energy.

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2 years ago

What are the units for gravitational field strength ?

Jobisdone [24]

Answer:

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

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3 years ago

Read 2 more answers

Energy is inversely proportional to the wavelength of a wave. Which would have the GREATEST energy? A wave with a

Ivan

Answer:

A 5

Explanation:

The wave with the least amount of wavelength will have the greatest amount of energy.

Wavelength and energy shares an inverse relationship;

E = h f = $\frac{hc}{wavelength}$

From this equation, we see that the higher the energy of a wave, the lesser its wavelength.

Choice A from the options has the least wavelength.
Wavelength is the distance between two successive crests of a wave.

This is why we see that in the electromagnetic spectrum, radio waves have the least energy because they have the longest wavelength.

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2 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

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3 years ago

I NEED HELP WITH THE LAST QUESTION I’L MARK U AS BRANLIEST

velikii [3]

Answer:

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