First question: 800J
Second question: 20.4m
I think in parallel circuits.
Answer:
0.72 Hz minimum frequency
Explanation:
When the damping is negligible,Amplitude is given as

here
= (6.30)/(0.135) = 46.67 N/m kg
= 1.70/(0.135)(0.480) = 26.2 N/m kg
From the above equation , rearranging for ω,

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47
⇒ ω = 8.53 or 4.52 rad/s
Frequency = f
ω=2 π f
⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz
The lower frequency is 0.72 Hz and higher is 1.36 Hz
Answer:
124.86 V
Explanation:
We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft
1 ft = 0.3048 m
358 ft = 109.12 m
The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m
The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²
Resistivity of wire (ρ) = 0.0171 Ω.mm²/m
The resistance of the wire = 
The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V
The voltage at end = 125.2 - 0.34 = 124.86 V