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:
19.53 cm
Explanation:
The computation of the height is as follows:
Here we applied the conservation of the energy formula
As we know that
P.E of the block = P.E of the spring
m g h = ( 1 ÷ 2) k x^2
where
m = 0.15
g = 9.81
k = 420
x = 0.037
So now put the values to the above formula
(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2
1.4715 (h) = 0.28749
h = 0.19537 m
= 19.53 cm
Answer:
Explanation:
Using the pythagoras theorem, the displacement is expressed as;
d² = x²+y²
y = 36m (north)
x = 20m east
Substitute;
d² = 36²+20²
d² = 1296+400
d² = 1696
d = √1696
d = 41.18m
For the direction;
theta = tan^-1(y/x)
theta = tan^-1(36/20)
theta = tan^-1(1.8)
theta = 60.95°
Hence the magnitude is 41.18m and the direction is 60.95°