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3 years ago

The planet earth orbits around the sun and also spins around its own axis. True or False

Physics

1 answer:

posledela3 years ago

5 0

Answer:

This is True

Explanation: The Earth rotates around its own axis, which results in day changing to night and back again. The Earth actually revolves around, or orbits, the sun. One revolution around the sun takes the Earth about 365 days, or one year. Forces at work in the solar system keep the Earth, as well as the other planets, locked into predictable orbits around the sun.

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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:

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 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

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7 0

3 years ago

Pls help with all questions dew in 10 minutes!

Artyom0805 [142]

7.Jupiter is the largest planet in our solar system at nearly 11 times the size of Earth and 317 times its mass.

When we look at Jupiter, we're actually seeing the outermost layer of its clouds.

The Great Red Spot is a storm in Jupiter's southern hemisphere with crimson-colored clouds that spin counterclockwise at wind speeds

8. 58,232 km

The second largest planet in the solar system

Surface. As a gas giant, Saturn doesn't have a true surface. The planet is mostly swirling gases and liquids deeper down.

Saturn's rings are thought to be pieces of comets, asteroids or shattered moons that broke up before they reached the planet,

9. Unlike the other planets of the solar system, Uranus is tilted so far that it essentially orbits the sun on its side, with the axis of its spin nearly pointing at the star.

Uranus' atmosphere is mostly hydrogen and helium, with a small amount of methane and traces of water and ammonia.

As an ice giant, Uranus doesn't have a true surface. The planet is mostly swirling fluids. While a spacecraft would have nowhere to land on Uranus, it wouldn't be able to fly through its atmosphere unscathed either. The extreme pressures and temperatures would destroy a metal spacecraft.

10. 24,622 km

Neptune has an average temperature of -353 Fahrenheit (-214 Celsius).

Neptune's atmosphere is made up mostly of hydrogen and helium with just a little bit of methane.

5 0

3 years ago

Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we

Scilla [17]

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>

Mass 1 = m₁
Mass 2 = m₂
Gravitational constant = G
Distance apart = r
Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>

Mass 1 = 2m₁
Mass 2 = 3m₂
Gravitational constant = G
Distance apart (r) = r
New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

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6 0

2 years ago

When catching a baseball, the ball applies a force of 39.6 N to a catcher's glove. If the work done on the catchers glove is 4

cricket20 [7]

Answer:

d = 1.19 m

Explanation:

Given that,

The force applied by the ball, F = 39.6 N

The work done on the catchers glove is 47.5 J

We need to find the distance traveled by the ball. We know that,

Work done, W = Fd

Where

d is the distance traveled

$d=\dfrac{W}{F}\\\\d=\dfrac{47.5 }{39.6 }\\\\d=1.19\ m$

So, it will cover 1.19 m.

4 0

3 years ago

Two positive charges are labeled q Subscript 1 baseline and q Subscript 2 baseline are 0.1 m apart. Two positively charged parti

Snowcat [4.5K]

Answer:

Explanation:

Let that point be at a distance x from q1

Then Kq1/x^2= Kq2/ (s-x)^2

Taking square roots and simplifying, x =s /[1+(q2/q1)^0.5]

Assuming an identical distance, the rigidity of Q on 2Q is equivalent in value to the rigidity of 2Q on Q. for that reason, had the area R been stored an identical, the two forces could be equivalent. inspite of the shown fact that, via fact the area is being decreased, we could constantly consult with the equation we use to calculate those forces: F = ok(Q1xQ2)/(R^2) because R is squared and is being halved, the final result's that's it being divided by potential of a million/4. for that reason, the rigidity would be expanded by potential of four, and be 4F.

4 0

3 years ago