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Art [367]
3 years ago
5

What did the protoplanets become?

Physics
2 answers:
alexira [117]3 years ago
7 0

What did the protoplanets become?

a. nebulae
b. planets
c. solar nebulae
d. planetesimals

The protoplanets become nebulae. The answer is letter A. The rest of the choices do not answer the question above.

nataly862011 [7]3 years ago
7 0

The answer is B. Planets

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What happens to the resistance of a wire as it gets wider
Alchen [17]

Answer:

it snaps

Explanation:

the more force you put on it, the wired out it gets than it snaps. I think

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3 years ago
How do Air molecules enable sound to travel from a radios speaker to your ears?
MatroZZZ [7]
The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.
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4 years ago
Read 2 more answers
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

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

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
Determine the acceleration due to gravity?
MissTica
That's a weird graph, but judging from the units the acceleration is the slope of the graph.

a = (0.8 - 0.3)/(0.16 - 0.055) = 4.76 m/s²
8 0
3 years ago
Mr Johnson launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how much time does it take
vazorg [7]

Answer:c

Explanation:

3 0
3 years ago
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