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2 years ago

If the force of the sun and the earth is 3.5×10 to the power of 22 N, how far apart are they?

Physics

1 answer:

Jobisdone [24]2 years ago

5 0

Gravitational force between 2 objects . . .

F = G · m₁ · m₂ / D²

-- You said that F = 3.5 x 10²² Newtons.

-- G = the gravitational constant = 6.67 x 10⁻¹¹ N m² / kg²

-- You want to find D .

F = G · m₁ · m₂ / D²

Multiply each side by D² . . . D² · F = G · m₁ · m₂

Divide each side by F . . . D² = G · m₁ · m₂ / F

So finally . . . D = √(G · m₁ · m₂ / F )

D = √(6.67 x 10⁻¹¹ N·m²/kg² · Earth mass · Sun mass / 3.5 x 10²² N)

<em>D = 4.37 x 10⁻¹⁷ · √(Earth mass · Sun mass) </em> meters

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Two glass marbles (1 and 2), each supported by a nylon thread, are rubbed against a piece of silk and then are placed near a thi

iragen [17]

Answer:

d.) Marbles 1 and 2 repel each other, but no interaction occurs with marble

Explanation:

Rubbing the glass against the silk charges the glass positively.

Assuming polarization is ignored,

●Glass won't conduct electrical charge. Therefore, acting as a insulator.

●neutral objects won't distribute it's charge when affected by charge objects in proximity. Therefore, over the marble surface, the charge remain spherically symmetrical (undisturbed)

Marble 1 is positive

Marble 2 is positive

Since marble 1 and 2 are positive, repulsion will occur.

marble 3 is neutral (still assuming polarization is ignored), therefore, no interaction with marble 1 and 2

D is the answer.

4 0

3 years ago

An elevator ascends at a constant speed of 4 m/s, how much time is required for the elevator in order to travel 120 m upwards?

vlabodo [156]

We simply use the formula,

$velocity =\frac{distance}{time}$

Given, velocity = 4 m per s and distance = 120 m.

Substituting these values, we get

$4\ m/s =\frac{120\ m}{time}\\\\\ time=\frac{120\ m}{4\ m/s} = 30\ s$ .

Thus, required time for the elevator in order to travel 120 m upwards is 30 s.

3 0

3 years ago

How much time does it take the cheetah to travel 500 meters, if its average speed is 70 meters per second?

Whitepunk [10]

Answer: 7.14

Explanation:

6 0

2 years ago

g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$