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3 years ago

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The Moon is attracted to the Earth. The mass of the Earth is 6.0x1024 kg and the mass of the Moon is 7.4x1022 kg. If the Earth a

nd Moon are 345,000 km apart, what is the gravitational force between them?

1 answer:

xxTIMURxx [149]3 years ago

7 0

Sorry I don't know the answer

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Skid Marx set the world record for acceleration as he went from zero to 10 m/s in 3 seconds. How far did he travel in that time?

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12000 miles per hour

Explanation:

If skid Marx went from zero to 10 miles in 3 seconds the his speed would have been 12000 miles per hour.

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Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the charge of object 1 is doubled AND the charge

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Explanation:

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Cubes have sides of equal lengths. What is the fastest way to measure the volume of a cube-shaped table?

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3 years ago

Choose all of the true statements regarding the relationship between voltage, resistance, and current. Current is measured in am

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-- <u><em>Current is measured in amps.</em></u> (You can use any symbol you want to represent current, but the most common one is " I ", not "Δ".)

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(When solving for Resistance in a circuit and both voltage and current are known values, the equation I =V*R is not true, and not the way to solve it.) <u><em>If the resistance is what you're looking for, then the equation to use is </em></u><u><em>R = V / I</em></u><u><em> . </em></u>

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3 years ago

Read 2 more answers

Two equally charged insulating balls each weigh 0.16 g and hang from a common point by identical threads 35 cm long. The balls r

Dafna11 [192]

Answer:

Q₁ = Q₂ = 8.84 x 10⁻⁹ C

Explanation:

given,

mass of ball, m = 0.16 g = 1.6 x 10⁻⁴ Kg

ball each other, r = 6.8 cm

Weight of the ball

F_w = m g

F_w = 1.6 x 10⁻⁴ x 9.8

F_w = 1.56 x 10⁻³ N

The tension in each string is a force directed along the length of the string and is the hypotenuse of a right triangle.

we have to find the horizontal component of the forces.

The length of the string,L is 35 cm so, it will be the hypotenuse.

θ be the angle made with imaginary vertical line and the string.

now,

$sin \theta = \dfrac{r\2}{L}$

$sin \theta = \dfrac{3.4}{35}$

θ = 5.57°

horizontal component of the force = ?

vertical component of force,F_v = 1.56 x 10⁻³ N

$tan\theta = \dfrac{F_H}{F_v}$

$tan(5.57^0) = \dfrac{F_H}{1.56\times 10^{-3}}$

F_h = 1.52 x 10⁻⁴ N

now, each ball will be repelled by

F = 1.52 x 10⁻⁴ N

now calculation of charges

$F = \dfrac{kQ_1Q_2}{r^2}$

Q₁ = Q₂ because both charge are same

$1.52\times 10^{-4} = \dfrac{9\times10^9Q^2}{0.068^2}$

Q² = 7.809 x 10⁻¹⁷

Q = 8.84 x 10⁻⁹ C

hence the change on the balls were Q₁ = Q₂ = 8.84 x 10⁻⁹ C

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3 years ago