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Irina-Kira [14]
3 years ago
13

The distance from Earth to the North Star is about 430 light-years. Which of the following statements describes why scientists u

se the light-year unit to measure distance in space?
A.) A light-year is very small and distance in space are very small

B.) A light-year unit is very small and the distance in space are very large

C.) A light-year unit is very large and distance in space are very small

D.) A light-year unit is very large and distance in space are very large
Physics
1 answer:
vampirchik [111]3 years ago
7 0

Answer:

D

Explanation:

D

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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
Miswer
quester [9]

Answer: 4.0

Explanation:

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2 years ago
The moon's surface gravity is one-sixth that of the earth. Calculate the weight on the moon of an object that has a mass of 24 k
ad-work [718]
When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
we mean that the acceleration of gravity on the Moon's surface is 1/6 of
the acceleration of gravity on the Earth's surface.

The acceleration of gravity is (9.8 m/s</span>²) on the Earth's surface, so
<span>it would be (9.8/6 m/s</span>²) on the Moon's surface.
<span>
The weight of any object, right now, is

(object's mass) </span>· (acceleration of gravity where the object is located now) .
<span>
If the object's mass is 24 kg and the object is on the Moon right now,
then its weight is 

(24 kg) </span>· (9.8/6 m/s²)

= (24 · 9.8 / 6) kg-m/s²

= 39.2 Newtons
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2 years ago
Which best describes what occurs when an object takes in a wave as the wave hits it?
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Answer: B. Absorption
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3 years ago
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A box has sides of 10 cm, 8.2 cm, and 3.5 cm. What is its volume?
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The volume would be 287cm³. Multiply all the 3 numbers by each other
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