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Rasek [7]
2 years ago
10

1.) A negative charge of -2.0x10-4 C and a positive charge of 8.0x10-4 C are

Physics
1 answer:
exis [7]2 years ago
6 0

Answer:

The Force between the two charges is an attractive force of 16,000N

Explanation:

Expression for the electric force between the two charges is given by

F = (k*q1*q2) / r^2

Here, k = constant = 9 x 10^9 N*m^2 / C^2

q1 = - 2.0x10^-4C

q2 = + 8.0x10^-4C

r = 0.30 m

Substitute the given values in the above expression -

One charge  is + and the other is a -, therefore the net force is an attractive force (opposites atract)

The attraction force is:

F= 9.0x10^9 * 2.0x10^-4 *8.0x10^-4 N/ 0.30^2

F= 16,000N

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Answer:

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2 years ago
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lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

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3 years ago
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