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Helga [31]
3 years ago
7

when drawing electric field lines ___________ charges have vectors point away/out and______ charges have vectors point toward/in

.
Physics
1 answer:
cupoosta [38]3 years ago
7 0

Answer:

Positive, Negative

Explanation:

The image I've attached shows that vectors point into the negative source and vectors point away from the positive source.

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68 rations * 0.1 dekarations/1 ration = 6.8 dekarations
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10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr
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Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

F_{g0}=G \frac{mM_{E0}}{r^{2}}

So if the mass of the earth is increased by a factor of 2, this means that:

M_{Ef}=2M_{E0}

so:

F_{gf}=G \frac{2mM_{E0}}{r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=2

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

M_{Ef}=3M_{E0}

so:

F_{gf}=G \frac{3mM_{E0}}{r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=3

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

M_{Ef}=\frac{M_{E0}}{4}

so:

F_{gf}=G \frac{mM_{E0}}{4r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=\frac{1}{4}

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0
3 years ago
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