Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz
Answer:

Explanation:
When a spring is compressed, the force exerted by the spring is given by:

where
k is the spring constant
x is the compression of the spring
In this problem we have:
k = 52 N/m is the spring constant
x = 43 cm = 0.43 m is the compression
Therefore, the force exerted by the spring on the dart is

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

where
F = 22.4 N is the force exerted on the dart by the spring
m = 75 g = 0.075 kg is the mass of the dart
a is its acceleration
Solving for a,

If they become closer, it is increased, and if the objects become farther away is decreased.
86.3 I just did the math and that’s the answer I got