Amber asked her roommate to turn down the radio because she was trying to study. Her roommate had increased the volume from a vo
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For an AC circuit:
I = V/Z
V = AC source voltage, I = total AC current, Z = total impedance
Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.
For a resistor R, inductor L, and capacitor C, their impedances are given by:
= R
R = resistance
= jωL
ω = voltage source angular frequency, L = inductance
= -j/(ωC)
ω = voltage source angular frequency, C = capacitance
Given values:
R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s
Plug in and calculate the impedances:
= 217Ω
= j(220)(0.875) = j192.5Ω
= -j/(220×6.75×10⁻⁶) = -j673.4Ω
Add up the impedances to get the total impedance Z, then convert Z to polar form:
Z = +
+
Z = 217 + j192.5 - j673.4
Z = (217-j480.9)Ω
Z = (527.6∠-1.147)Ω
Back to I = V/Z
Given values:
V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)
Z = (527.6∠-1.147)Ω (from previous computation)
Plug in and solve for I:
I = (30.0∠0+220t)/(527.6∠-1.147)
I = (0.0569∠1.147+220t)A
To get the voltages of each individual component, we'll just multiply I and each of their impedances:
= I×
= I×
= I×
Given values:
I = (0.0569∠1.147+220t)A
= 217Ω = (217∠0)Ω
= j192.5Ω = (192.5∠π/2)Ω
= -j673.4Ω = (673.4∠-π/2)Ω
Plug in and calculate each component's voltage:
= (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V
= (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V
= (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V
Now we have the total and individual voltages as functions of time:
V = (30.0∠0+220t)V
= (12.35∠1.147+220t)V
= (10.95∠2.718+220t)V
= (38.32∠-0.4238+220t)V
Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:
V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V
= 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V
= 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
= 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V
Answer:
2.7 s
Explanation:
The period of the motion of a massless loaded spring is given by
where m is the mass of the load and k is the force or spring constant.
Using values in the question and converting to appropriate units,
Answer:
C) solo III
Explanation:
Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.
I) el cuerpo pesa igual que su masa.
Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.
donde:
w = peso [N]
m = masa [kg]
g = aceleracion gravitacional = 9.81 [m/s²]
Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.
II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.
III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.
Esta es la opcion valida, la fuerza neta es nula.