Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?
2 answers:
<h2>Question :- </h2>
Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?
<h2>Answer :-</h2><h3>Gravitational Force :- </h3>- It is the attractive force between the two body having some mass.
- Example of gravitational force :- attraction between the human and earth . earth pulls the body towards it with the help of this
- This force helps the celestial bodies to revolve in there orbits
- Mass :- gravitational force directly proportional on the masses of the body between which it is acting
- Distance between the masses :- Gravitational force is inversely proportional to the square of the distance between the masses
where
- F is gravitational force
- G is gravitational constant ( universal constant )
- m 1 is mass of first body
- m2 is mass of second body
- r is the distance between the two bodies
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Answer:
(a) The potential difference between the spheres is 750 kVA
(b) The charge on the smaller sphere is 6. μC
(c) The charge on the smaller sphere, Q₁ = 13. μC
Explanation:
(a) The given parameters are;
The charge on the inner sphere, q = 5.00 μC
The radius of the inner sphere, r = 3.00 cm = 0.03 m
The charge on the larger sphere, Q = 15.0 μμC
The radius of the larger sphere, R = 6.00 cm = 0.06 m
The potential difference between two concentric spheres is given according to the following equation;
Where;
R = The radius of the larger sphere = 0.06 m
r = The radius of the inner sphere = 0.03 m
q = The charge of the inner sphere = 5.00 × 10⁻⁶ C
Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C
k = 9 × 10⁹ N·m²/C²
Therefore, by plugging in the value of the variables, we have;
The potential difference between the spheres, = 750,000 N·m/C = 750 kVA
(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;
= Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C
= 20 × 10⁻⁶ C
From which we have;
Q₁/Q₂ = R/r
Where;
Q₁ = The new charge on the on the larger sphere
Q₂ = The new charge on the on the smaller sphere
= 20 × 10⁻⁶ C = Q₁ + Q₂
∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂
∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2
20 μC - Q₂ = 2·Q₂
20 μC = 3·Q₂
Q₂ = 20 μC/3
The charge on the smaller sphere, Q₂ = 20 μC/3 = 6. μC
(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3
The charge on the smaller sphere, Q₁ = 40 μC/3 = 13. μC.
According to Newton's third law
- Ever action has a equal and opposite reaction.
So
- Hence if applied force is 8000N opposing force is also 8000N
As car is pushed forward so the opposing force is less than 8000N
The final speed of C is equal to that of B.
The travel time of A is less than that of B.
The final speed of B is equal to that of A.
The travel time of C is greater than that of B.
The final speed of A is equal to that of C.
The travel time of C is greater than that of A.
<h3>What is speed?</h3>The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.
Consider the three inclines shown in the figure (attached below). The vertical line is an incline with a 90° angle. All three inclines are frictionless. Three small identical objects are released from the top of the inclines.
When the objects are at the same height, they travel with the same height. The travel time is directly proportional to the incline.
Thus, all the statements are completed.
Learn more about speed.
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