A free-falling object is an object moving under the effect of gravitational forces alone
The correct option to select for the True or False question is False
The reason the above selected option is correct is as follows:
According to Newton's second law of motion, we have;
Force = Mass × Acceleration
The force of gravity is 
Where;
m = The mass of the object
∴ The force acting on an object in free fall, 
= m × g
Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time
The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>
<u></u>
Learn more about object in free fall here:
brainly.com/question/13712424
brainly.com/question/11698474
Explanation:
In local galactic group the force of expansion of universe is overcome by the force of attraction due to gravity. Best example is our own galaxy milky way and another giant galaxy in our local group Andromeda. Andromeda having enormous gravity is pulling milky way towards itself, overcoming the force of expansion.
So, there are possibilities of collision despite the expansion of universe at a rapid pace. It is estimated that the milky way and Andromeda will collide each other after about 50 billion years from now.
75000 lol enjoy..............using up 20 characters
<h2>
Answer:</h2>
In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:
So the average power is:
But:
So:
![P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%28%5Cfrac%7B1%2Bcos%282%5Comega%20t%29%7D%7B2%7D%20%29dt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7Bcos%282%5Comega%20t%29%7D%7B2%7D%5Ddt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B4%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2T%7D%5B%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B2%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2%7D)
In terms of RMS values:
