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3 years ago

A point charge of 6.30 uC is lecated 15.0 cm from a second point charge of-4.80 yC. The force on the 6.30 juC charge is

Physics

1 answer:

faltersainse [42]3 years ago

7 0

Answer:

12.096 N

Explanation:

Q = 6.30 micro coulomb = 6.3 x 10^-6 C

q = - 4.8 micro coulomb = - 4.8 x 10^-6 C

d = 15 cm = 0.15 m

According to the Coulomb's law, the force between the two charges is given by

$F = \frac{KQq}{d^{2}}$

Where, k be the Coulombic constant which is equal to 9 x 10^9 Nm^2 / C^2. This value is for free space or vacuum be the medium between two charges.

By substituting the values

$F = \frac{9\times 10^{9}\times 6.3 \times10^{-6}\times 4.8\times 10^{-6}}{0.15\times 0.15}$

F = 12.096 N

This force is attractive in nature as the charges as unlike in nature and by the properties of charges, if there are two unlike charges, then the force between the charges is attractive in nature.

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4 years ago

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The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU

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Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

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3 years ago

When the potential energy is high, the kinetic energy is _____.

kramer

Answer:

low then it gets high

Explanation:

none

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4 years ago

A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and th

gizmo_the_mogwai [7]

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

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3 years ago

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a

s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = $0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg$

r = Radius of orbit of planet = $0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m$

v = Orbital speed

The kinetic and potential energy balance is given by

$\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s$

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

$t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days$

The orbital period is 14.45 days

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3 years ago