16.5375 .................
Answer
given,
capacitance = C = 3.4-µF
inductance = L = 0.08 H
frequency is expressed as
![f = \dfrac{1}{2\pi\sqrt{LC}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1%7D%7B2%5Cpi%5Csqrt%7BLC%7D%7D)
time period
![T = \dfrac{1}{f}=2\pi\sqrt{LC}](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7Bf%7D%3D2%5Cpi%5Csqrt%7BLC%7D)
after time T/4 current reach maximum
![t = \dfrac{T}{4}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7BT%7D%7B4%7D)
![t = \dfrac{2\pi\sqrt{LC}}{4}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B2%5Cpi%5Csqrt%7BLC%7D%7D%7B4%7D)
![t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B2%5Cpi%5Csqrt%7B0.08%20%5Ctimes%203.4%20%5Ctimes%2010%5E%7B-6%7D%7D%7D%7B4%7D)
t = 8.2 x 10⁻⁴ s
t = 0.82 ms
b) using law of conservation
![\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7DCV%5E2%3D%5Cdfrac%7B1%7D%7B2%7DLI%5E2)
![I^2 = \dfrac{CV^2}{L}](https://tex.z-dn.net/?f=I%5E2%20%3D%20%5Cdfrac%7BCV%5E2%7D%7BL%7D)
![I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}](https://tex.z-dn.net/?f=I%5E2%20%3D%20%5Cdfrac%7BC%7D%7BL%7D%5Cdfrac%7BQ%5E2%7D%7BC%5E2%7D)
![I =\sqrt{\dfrac{Q^2}{CL}}](https://tex.z-dn.net/?f=I%20%3D%5Csqrt%7B%5Cdfrac%7BQ%5E2%7D%7BCL%7D%7D)
![I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}](https://tex.z-dn.net/?f=I%20%3D%5Csqrt%7B%5Cdfrac%7B%285.4%20%5Ctimes%2010%5E%7B-6%7D%29%5E2%7D%7B0.08%20%5Ctimes%203.4%20%5Ctimes%2010%5E%7B-6%7D%7D%7D)
I = 0.010 A
I = 10 mA
Decreases the input force
The period T of a pendulum is given by:
![T=2 \pi \sqrt{ \frac{L}{g} }](https://tex.z-dn.net/?f=T%3D2%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg%7D%20%7D%20)
where L is the length of the pendulum while
![g=9.81 m/s^2](https://tex.z-dn.net/?f=g%3D9.81%20m%2Fs%5E2)
is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is
![T=0.200 s](https://tex.z-dn.net/?f=T%3D0.200%20s)
. Using this data, we can solve the previous formula to find L:
At position of maximum height we know that the vertical component of its velocity will become zero
so the object will have only horizontal component of velocity
so at that instant the motion of object is along x direction
while if we check the acceleration of object then it is due to gravity
so the acceleration of object is vertically downwards
so it is along y axis
so here these two physical quantities are perpendicular to each other
so correct answer would be
<em>C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other. </em>