the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force bet
2 answers:
The gravitational force between the objects A. It would increase.
Explanation:
The magnitude of the gravitational force between two objects is given by:
where
G is the gravitational constant
are the masses of the two objects
r is the separation between the objects
In this problem, we are told that one of the object (the one on the right) gains mass: this means that, for instance, the value of increases. We can see from the equation that the gravitational force is directly proportional to the masses: therefore, if one of the masses increases (while the distance between the two objects remains constant), it means that the force also increases.
Therefore, the correct answer is
A. It would increase.
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Answer:
Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
<u><em>Now,</em></u>
From the question,
V(t) = 230 sin (100t) ---------------(ii)
<em><u>By comparing equations (i) and (ii) the following holds;</u></em>
V = 230
ω = 100
Ф = 0
<em><u>But;</u></em>
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = Hz
Therefore, the frequency of the voltage is Hz
Answer:
Height, h = 10.20 meters
Explanation:
It is given that,
Mass of the object, m = 3 kg
Energy of object, E = 300 J
Let it will moved to a height of h. The energy possessed by it is called gravitational potential energy. It is given by :
h = 10.20 meters
So, the object will move to height of 10.20 meters. Hence, this is the required solution.
Answer:
Acceleration,
Explanation:
It is given that,
Time period of revolution of the moon,
If the distance from the center of the moon to the surface of the planet is,
The radius of the planet,
Let a is the moon's radial acceleration. Mathematically, it is given by :
, R is the radius of orbit
Since,
The radius of orbit is,
So,
Hence, this is the required solution for the radial acceleration of the moon.