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3 years ago

In a compressional wave the denser the medium is at the compressions the smaller is amplitude?

1 answer:

5 0

Wellll, let me think about that ...

I don't think I'd agree that you can characterize the amplitude of
a wave according to the density at only one point in it. After all ...
a tiny wave in steel would be much denser at a compression than
a huge wave in air would be.

The amplitude of any wave is described as the difference between
a peak and the resting value. Or even better ... half of the difference
between a maximum and a minimum.

So if you're looking at a longitudinal wave, like sound, I'd say if you
want to describe its amplitude, then you have to look at the density
at two points ... either the difference between the compression and
the resting densities, or the difference between the greatest compression
and the greatest rarefaction.

That's my opinion. I could be wrong.

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If a girl is standing in front of a smooth surface from which a sound is reflected, the girl may hear

gavmur [86]

Near the surface of reflection, reflected wave may interfere with incident wave leading to production of constructive and as well as destructive interference. This in turn, can result to resonance as well as enhancement of the sound intensity as the waves of reflection adds to incident wave. Therefore, the girl would higher intensity of reflected waves as compared to incident waves.

Therefore, statement A is correct.

3 0

3 years ago

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The official web site of the Nobel Prize explains that Marie Curie’s chemistry prize was partly for her discovery that the radio

anastassius [24]

The official web site of the Nobel Prize explains that Marie Curie’s chemistry prize was partly for her discovery that the radioactivity of a substance is unaffected when it undergoes a chemical reaction. The discovery implied was that, Radioactivity involves Radioactivity involves only neutrons.

Explanation:

The official web site of the Nobel Prize explains that Marie Curie’s chemistry prize was partly for her discovery that the radioactivity of a substance is unaffected when it undergoes a chemical reaction. The discovery implied was that, Radioactivity involves only neutrons.
Marie Curie studied about the radiation of all compounds containing the known radioactive elements, including uranium and thorium, which she later discovered that they were radioactive.
she discovered the following results,
the exact measurement of the strength of the radiation from uranium;
the intensity of the radiation was found to be proportional to the amount of uranium or thorium in the compound .
the ability to emit radiation is not dependent on the arrangement of the atoms in a molecule;
it must be linked to the interior of the atom itself which is a revolutionary discovery.

8 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )