The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.
All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.
If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.
Answer: True
Explanation: Inductors are similar to resistors, due to the fact that they offer resistance to current flow, but Inductors are different from resistors in that, while resistors loss electric energy in a circuit in the form of heat, an inductor stores that energy in the form of a magnetic field.
As current passes through an inductor overtime it tends to store current in the form of magnetic field. Therefore the electric-power industry can store energy in large Inductors.
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
Substituting,
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
Where here 
is the velocity after the collision.
The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.
The time period T of the block with mass M attached to a spring of spring constant K is given by,
T = 2π(√M/K).
Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,
T' = 2π(√2M/K)
Putting T = 2π(√M/K) above,
T' =√2T
So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.
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