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lilavasa [31]
3 years ago
10

If a 50 KG object is at a location 25,600 km from Earth's Center, what is the gravitational force exerted by the objects on Eart

h? In what direction does that force act? Support your answer with evidence.
Physics
1 answer:
masya89 [10]3 years ago
5 0

Answer: The force exerted by the object on Earth is 30.4 N, and the direction is towards the object.

Explanation:

The gravitational force between two objects of mass M1 and M2, that are at a distance R between them is written as:

F = G*(M1*M2)/R^2

Where G is the gravitational constant, such that:

G = 6.674*1^(-11) m^3/(kg*s^2)

We know that M1, the mass of the object, is 50kg

R is the distance, in this case, is 25,600km

But this needs to be written in meters, remembering that:

1km = 1000m

Then:

25,600km = (25,600*1000)m = 25,600,000 m

And M2 is the mass of Earth, which is:

M2 = 5.972*10^(24) kg

Replacing all of those in the force equation we get

F = (6.674*1^(-11) m^3/(kg*s^2))*(50kg*5.972*10^(24) kg)/( 25,600,000 m)^2

F = 30.4 N

We know that the gravitational force is attractive, then the direction in which this force acts is towards the 50kg object.

Now, remember the second Newton's law is:

F = M*a

Then the acceleration that the object causes on Earth is:

30.4N = ( 5.972*10^(24) kg)*a

a = 30.4N/( ( 5.972*10^(24) kg)) = 5.09*10^(-24) m/s

This acceleration is almost despreciable.

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E = 2150 N/C

Part b)

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F = qE

F = 1.6 * 10^{-19} * 2150

F = 3.44 * 10^{-16} N

PART C)

Gravitational force is given by

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F_g = 1.6 * 10^{-27}*9.8

F_g = 1.57 * 10^{-26} N

PART d)

Ratio of electric force to weight

\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}

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Answer:

Option B

Explanation:

Speed of a wave is denoted by:

v=fλ

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v=fλ

v=15(28)\\v=420m/s

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Answer:

140°

Explanation:

The law of reflection states that the angle of redlection equals to the angle of incidence.

When light rays hit surface at 20°, they also leave the surface at the same angle

Since the whole surface has 180° then subtracting these two angles from total angle gives the the angle between the incident and reflected rays.

180°-20°-20°=140°

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-v/2

Explanation:

Given that:

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  • Collides with the wall going through a sliding motion on on the plane smooth surface.
  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

KE_i=\frac{1}{2}. m.v^2

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

KE_f=\frac{1}{4}\times KE_i

KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2

Considering that the mass of the body remains constant before and after collision.

KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

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