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melomori [17]
3 years ago
7

A force of 5.25 newtons acts on an object of unknown mass at a distance of 6.9 × 108 meters from the center of Earth. To increas

e the force to 2.5 times its original value, how far should the object be from the center of Earth?
Physics
2 answers:
alexira [117]3 years ago
7 0

Answer:

4.36 x 10^8 m

Explanation:

Let the mass of earth is M and unknown mass is m.

According to the Newton's law of gravitation, force between two objects is given by

F = G\frac{M m}{d^{2}}

Here, F = 5.25 N, d = 6.9 x 10^8 m

5.25 = G\frac{M m}{(6.9\times 10^{8})^{2}}       .... (1)

Now, F' = 2.5 F and d be the distance

2.5\times 5.25 = G\frac{M m}{(d)^{2}}            ..... (2)

Divide equation (1) by equation by (2)

\frac{1}{2.5} = \left ( \frac{d}{6.9\times 10^{8}} \right )^{2}

d = 4.36 x 10^8 m

NemiM [27]3 years ago
5 0
The force between two objects is calculated through the equation, 
                        F = Gm₁m₂/d²
where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is universal gravitation constant. 

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,
                       2.5F = Gm₁m₂ / (D²)
                                D = 0.623d

Subsituting the known value of d,
                                D = 0.623(6.9 x 10^8) = <em>4.298 x 10^8 m</em>
                           
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Explanation:

Given that,

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Force acting on the object, F=19\ nC=19\times 10^{-9}\ C (in downward direction)

(a) The electric force acting in the electric field is given by :

F=qE

E is the electric field

E=\dfrac{F}{q}

E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, q=1.6\times 10^{-19}\ C

The force acting on the proton is :

F=qE

F=1.6\times 10^{-19}\times 4.75

F=7.6\times 10^{-19}\ N

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0
3 years ago
The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

g \approx 9.82\,\frac{m}{s^{2}}

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

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ANYTHING you drop does that, if air resistance doesn't hold it back.

7 0
3 years ago
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