The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.
The given parameters;
- time of fall of the first ball, t = 1 s
- time of fall of the second ball, t = 3 s
The distance traveled by each ball is calculated using the second equation of motion as shown below.
The distance traveled by the first ball is calculated as follows;
![h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m](https://tex.z-dn.net/?f=h%20%3D%20u_0t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%20%280.5%5Ctimes%209.8%5Ctimes%201%5E2%29%5C%5C%5C%5Ch%20%3D%204.9%20%5C%20m)
The distance traveled by the second ball is calculated as follows;
![h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%20%280.5%5Ctimes%209.8%5Ctimes%203%5E2%29%5C%5C%5C%5Ch%20%3D%2044.1%5C%20m)
Thus, the second ball traveled a greater distance because it spent more time in motion.
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The total circuit current at the resonant frequency is 0.61 amps
What is a LC Circuit?
- A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
- These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.
Q =15 = (wL)/R
wL = 30 ohms = Xl
R = 2 ohms
Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance
| Zs | = 30.07 <86.2° ohms
Xc = 1/(wC) = 30 ohms
The impedance of the LC circuit is found from:
Zp = (Zs)(-jXc)/( Zs -jXc)
Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°
I capacitor = 277/-j30 = j9.23 amps
I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps
I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps
Hence, the total circuit current at the resonant frequency is 0.61 amps
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