The formula is
F_grav = G * m1 * m2 / r^2
G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.
As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.
As the distance decreases, the gravitational force will Increase.
The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.
Answer:
9000RPM
Explanation:
"Angular velocity" is directly related to kinetic energy, that is, the Kinetic energy equation would allow an approximation to the resolution investigated in the problem.
The equation for KE is given by:

Now, starting from there towards the <em>Angular equation of kinetic energy</em>, the moment of inertia (i) is used instead of mass (m), and angular velocity (w) instead of linear velocity (V)
That's how we get

calculating the inertia for a solid cylindrical disk, of
m = 400kg
r = 1.2 / 2 = 0.6m

We understand that the total kinetic energy is 3.2 * 10 ^ 7J, like this:



Thus,
943 rad / s ≈ 9000 rpm
The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.
<u>Explanation:</u>
A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:

Here L is the inductance and I is the current.
Here, energy stored (U) is proportional to the number of turns (N) and the current (I).

mu not - permeability of core material
A -area of cross section
l - length
N - no. of turns in solenoid inductor
Now,given that the proportion always remains same:

In this way the expression


Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.
a is the correct
Explanation:
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If a mass m is attached to an ideal massless spring and has a period of t, then the period of the system when the mass is 2m is
.
Calculation:
Step-1:
It is given that a mass m is attached to an ideal massless spring and the period of the system is t. It is required to find the period when the mass is doubled.
The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T.
It is known that the period is calculated as,

Here m is the mass of the object, and k is the spring constant.
Step-2:
Thus the period of the system with the first mass is,

The period of the system with the second mass is,

Then the period of the system with the second mass is
times more than the period of the system with the first mass.
Learn more about period of a spring-mass system here,
brainly.com/question/16077243
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