Answer:
Here the source is moving away from the observer so frequency will be smaller than the actual frequency and since the speed is increasing so the frequency is decreasing with time so correct answer is
D) lower than the original pitch and decreasing as he falls.
Explanation:
As we know by the Doppler's effect of sound we have
so we will have

so here when source moves away from the observer with a some speed then the frequency of the sound observed by the observer is smaller than the actual frequency
Here we know that the speed of the source is increasing with time as the source is falling under gravity
So we can say that the pitch of the sound will decrease with time
They are unbalanced forces ..... Hope this helps :3
I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it. But calculating the moment is easy, and we
can get along without the drawing.
Each separate weight has a 'moment'.
The moment of each weight is:
(the weight of it) x (its distance from the pivot/fulcrum) .
That's all there is to a 'moment'.
The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).
That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do. The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.
The answer is 9.8 ms^-2, because there is only one force acting on the object so the acceleration will be numerically equal to the gravitational field strength.
Answer:
Part a)

Part b)

Explanation:
Part a)
As we know that orbital velocity at certain height from the surface of Earth is given as

here we know that



now we have


Part b)
When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as
