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Lady bird [3.3K]
3 years ago
7

The granulation pattern that astronomers have observed on the surface of the Sun tells us that: a. the Sun is a lot cooler on th

e inside than on the outside b. the Sun's surface is made of a thin solid that cracks easily c. hot material must be rising from the Sun's hotter interior d. the solar wind must consist of very small (low-mass) particles e. the Sun accumulates a lot of dirt and dust because of its large gravity 5. Which statement about the Sun’s rotation is TRUE?
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

c. hot material must be rising from the Sun's hotter interior

Explanation:

Granulation is the grainy appearance of the solar photosphere produced by the top of the convection cells in the sun.

The grainy appearance are produced by granules on the photosphere of the sun and granules are caused by convection currents of plasma within the sun's convection zone.

The interior of these granules are brighter (and thus hotter) than the exterior of the granules which are darker.

<u>So, the granulation pattern that astronomers have observed on the surface of the Sun tells us that hot material must be rising from the Sun's hotter interior.</u>

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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:

1)

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})

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}

3)

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))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
Which ball (if either) has the greatest speed at the moment of impact
Ierofanga [76]
Are there any options??


I would have to say metal of course but without options I can't assume anything
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Look again at the pictures of the Marina District and Candlestick Park. They showed the very different effects of the same earth
lyudmila [28]

Explanation:

The deeper the sediment layer above bedrock, the more soft soil there is for the seismic waves to travel through. Soft soil means bigger waves and stronger amplification. The earthquake damage to this building may have been influenced by the type of soil it's sitting on.

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When the temperature of water increases from room temperature to 90*C the process of heating the water is
polet [3.4K]
The process is called ENDOTHERMIC
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If two waves interact in destructive interference, which of the following will be true regarding the new wave formed?
AlekseyPX

Answer:

b

Explanation:

6 0
3 years ago
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