Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by
![Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2](https://tex.z-dn.net/?f=Z%3D%5Csqrt%7BR%5E2%2B%5Cleft%20%28%20X_L-X_C%20%5C%7Dright%20%29%5E2)
but we know that at resonance
putting
in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
so from given expression it is clear that quality factor depends on R L and C
Answer:
The y-component of the electric force on this charge is ![F_y = -1.144\times 10^{-6}\ N.](https://tex.z-dn.net/?f=F_y%20%3D%20-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20N.)
Explanation:
<u>Given:</u>
- Electric field in the region,
![\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%28148.0%5C%20%5Chat%20i-110.0%5C%20%5Chat%20j%29%5C%20N%2FC.)
- Charge placed into the region,
![q = 10.4\ nC = 10.4\times 10^{-9}\ C.](https://tex.z-dn.net/?f=q%20%3D%2010.4%5C%20nC%20%3D%2010.4%5Ctimes%2010%5E%7B-9%7D%5C%20C.)
where,
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
![\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%5Cdfrac%7B%5Cvec%20F%7D%7Bq%7D%5C%5C%5Ctherefore%20%5Cvec%20F%20%3D%20q%5Cvec%20E%5C%5C%3D%2810.4%5Ctimes%2010%5E%7B-9%7D%29%5Ctimes%20%28148.0%5C%20%5Chat%20i-110.0%5C%20%5Chat%20j%29%5C%5C%3D%281.539%5Ctimes%2010%5E%7B-6%7D%5C%20%5Chat%20i-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20%5Chat%20j%29%5C%20N.)
Thus, the y-component of the electric force on this charge is ![F_y = -1.144\times 10^{-6}\ N.](https://tex.z-dn.net/?f=F_y%20%3D%20-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20N.)
Upstream speed = S - 1
Downstream speed = S + 1
Average speed = total distance / total time
Average speed = (S - 1) + (S + 1) / 2
= S
S = 6 miles / 4 hours
S = 1.5 miles per hour
Answer:
Ff = 19.6 N
Explanation:
So since its saying whats the minimum F to move the block, we will use static friction (0.5).
We will use the equation for force of friction, which is Ff = uFn
Ff = (0.5)(4)(9.8)
Ff = 19.6 N
this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block