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Alenkinab [10]
3 years ago
10

Two asteroids exert a gravitational force, F, on each other. Some time later, the asteroids are now three times as far from each

other as before. Which of the following represents the gravitational force at this distance? Select one:
a. F/9
b. F/6
c. F/3
d. F/2
Physics
1 answer:
dolphi86 [110]3 years ago
7 0

Answer:

a.F/9

Explanation:

Hello, I can help you wit this

the gravitational Force is described by the next equation

F=G*\frac{m_{1}*m_{2} }{r^{2} } \\

Where, G  is   the universal gravitational constant, M1 and M2 are the masses of the objects, and r is the distance between then.

Step 1

time 1

Let

m1= mass of the asteroid 1

m2= mass of the asteroid 2

distance 1=r1

put the values into the equation to find the force

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{1}=G*\frac{m_{1}*m_{2} }{r_{1}^{2} }

Step 2

time 2

r2=3r1

F_{2} =G*\frac{m_{1}*m_{2} }{(3*r_{1} )^{2} }\\

Step 3

compare the forces

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{2} =G*\frac{m_{1}*m_{2} }{9(r_{1}^{2}) }\\F_{2}=\frac{F_{1}}{9}

a.F/9

here it is observed that the force is inversely proportional to the square of the distance, at double distance, it is divided into 4, triple in 9, etc.

Have a great day.

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A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e
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Answer:

  V_inside = 36 V

Explanation:

<u>Given  </u>

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.  

<u>Required</u>

We are asked to calculate the potential at the centre of the sphere  

<u>Solution</u>

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V = (1/4*π*∈o) × (q/r)                                          (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V  

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside                               (2)

Now we can plug our values for r, R and V_outside into equation (2) to get  V_inside

V_inside = (1.2 m )/(0.600)*18

               = 36 V

  V_inside = 36 V

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