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3 years ago

11

What is one of the principles of charges in a conductor?

1 answer:

nevsk [136]3 years ago

3 0

Answer:

A conductor allows free charges to move about within it. The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. Any excess charge will collect along the surface of a conductor. Conductors with sharp corners or points will collect more charge at those points.

Explanation:

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What type of eclipse is shown? How do you know?

Sonja [21]

That is a lunar eclipse. At night, when the Earth is between the Sun and the moon, the moon would appear to be red. Just for future reference, a solar eclipse is when the Moon is between the Sun and Earth. Speaking of which, check out the solar eclipse this August!

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4 years ago

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Two coulombs of charge pass through a wire in 5 seconds. Calculate the current in the wire. 1

gogolik [260]

Answer:

0.4A.

Explanation:

Current (A) = Charge (coulomb)/Time (secs)

2 coulombs/5 secs = 0.4A

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3 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

Kipish [7]

Answer:

The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

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3 years ago

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

3 years ago