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exis [7]
2 years ago
12

What is the OUTPUT for this equation if the INPUT is 20? y = 2x

Mathematics
1 answer:
Alexeev081 [22]2 years ago
5 0
It should be 40







Mark be as brainliest if right plz
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You basically look at the bar graph on the right and see which statement is true. Brainliest and thank you!
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Step-by-step explanation:

To check this answer, you could obviously graph the points, but without graphing, you can tell the line would be straight because each x value is multiplied by the same number to get the corresponding y value.

3 x 5 = 15, 4 x 5 = 20, etc. This shows linear growth. If each x value were multiplied by twice itself (3 x 6, 4 x 8, and so on), you wouldn't have a straight line because the number that x is multiplied by changes depending on the value of x.

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Step-by-step explanation:

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A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and
natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: E=k\frac{q}{r^{2} }; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: p=\frac{q}{A}; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: S=4\pi r^{2}; which gives us the surface of a sphere.

The inner surface: Si=4\pi (0.06m)^{2} = 0.04 m^{2}

The outer surface: S=4\pi (0.08m)^{2}=0.08m^{2}

Now we can calculate the charges,

Inner charge: Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC

Outer charge: Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC

Then, we are able to calculate both fields:

Inner field: Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}

Outer field:  Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

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Hii plss helpp quick i have 3 minn
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