Answer:
Part a)
![f = 70.9 Hz](https://tex.z-dn.net/?f=f%20%3D%2070.9%20Hz)
Part b)
U = 111 J
Explanation:
As we know that the capacitor is of capacitance
![C = 24 mF](https://tex.z-dn.net/?f=C%20%3D%2024%20mF)
![V = 170 Volts](https://tex.z-dn.net/?f=V%20%3D%20170%20Volts)
now the maximum charge on it
![Q = CV](https://tex.z-dn.net/?f=Q%20%3D%20CV)
![Q = (24 mF)(170 V)](https://tex.z-dn.net/?f=Q%20%3D%20%2824%20mF%29%28170%20V%29)
![Q = 4.08 C](https://tex.z-dn.net/?f=Q%20%3D%204.08%20C)
Part a)
Oscillation frequency of the charge is given as
![f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7B1%7D%7BLC%7D%7D)
![f = 70.9 Hz](https://tex.z-dn.net/?f=f%20%3D%2070.9%20Hz)
Part b)
Now the equation of charge oscillation is given as
![q = Q cos(\omega t)](https://tex.z-dn.net/?f=q%20%3D%20Q%20cos%28%5Comega%20t%29)
![q = 4.08 cos(2\pi(70.9) t)](https://tex.z-dn.net/?f=q%20%3D%204.08%20cos%282%5Cpi%2870.9%29%20t%29)
now current in the circuit is given as
![i = \frac{dq}{dt}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7Bdq%7D%7Bdt%7D)
![i = (4.08)(2\pi(70.9)) sin(2\pi(70.9) t)](https://tex.z-dn.net/?f=i%20%3D%20%284.08%29%282%5Cpi%2870.9%29%29%20sin%282%5Cpi%2870.9%29%20t%29)
now at t = 1.35 ms we have
![i = 1028.36 A](https://tex.z-dn.net/?f=i%20%3D%201028.36%20A)
So the energy stored in inductor is given as
![U = \frac{1}{2}Li^2](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7B1%7D%7B2%7DLi%5E2)
![U = \frac{1}{2}(0.210 \times 10^{-3})(1028.36)^2](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.210%20%5Ctimes%2010%5E%7B-3%7D%29%281028.36%29%5E2)
![U = 111 J](https://tex.z-dn.net/?f=U%20%3D%20111%20J)
I believe it is, All of the above.
The acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).
<h3>Conservation of angular momentum</h3>
The acceleration of the mass down the plane is determined by applying the principle of conservation of angular momentum.
Fr = Iα
where;
- F is weight of the object parallel to the plane
- r is the radius of the flywheel
- I is moment of inertia
- α is angular acceleration
(mg sinθ)r = Iα
(mg sinθ)r = I(ar)
(mg sinθ) = I(a)
![a = \frac{mg \times sin(\theta)}{I} \\\\a = \frac{mg \times sin(\theta)}{3mr^2/4} \\\\a = \frac{4mg \times sin(\theta)}{3mr^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bmg%20%5Ctimes%20sin%28%5Ctheta%29%7D%7BI%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7Bmg%20%5Ctimes%20sin%28%5Ctheta%29%7D%7B3mr%5E2%2F4%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B4mg%20%5Ctimes%20sin%28%5Ctheta%29%7D%7B3mr%5E2%7D)
Thus, the acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).
Learn more about acceleration here: brainly.com/question/14344386
Downward force acting on the ball is 19.6N
Net force acting on the ball is 1960V N
<u>Explanation:</u>
<u />
Given:
Mass of the ball, m = 2kg
Density of ball, σ = 800 kg/m³
Density of water, ρ = 1000 kg/m³
Downward force acting by the ball in the vessel = mg
where, g = 9.8m/s²
F = 2 X 9.8
F = 19.6N
Net force acting on the ball:
Fnet = (ρ - σ) Vg
where,
V is the volume of water
Fnet = (1000 - 800) V X 9.8
Fnet = 1960V N
If the volume is known, then substitute the value of V to find the net force.
Thus, Downward force acting on the ball is 19.6N
Net force acting on the ball is 1960V N