1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..
And I hope this helped :)
The acceleration formula goes like this: a= (vf-vi)/t so it would be (13-4)/3 Thus the answer is 3m/s^2
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
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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:

Answer:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by

where
E is the electric field
A is the cross-sectional area
is the angle between the direction of the electric field and the normal to A
The flux is maximum when
, so we are in this situation and therefore
, so we can write

Here we have:
is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is

And so, we can find the magnitude of the electric field:
