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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
Explanation:
It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.
So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.
U = qV
Since, the electric potential is directly proportional to the charge. Hence, when 0.2 
tends to replaced by 0.4 then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.
Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.
Answer: conduction :it transfers heat between objects that are in direct contact with eachother
