Answer:
E) 6.5 A
Explanation:
Given that
L = 40 m H
C= 1.2 m F
Maximum charge on capacitor ,Q= 45 m C
The maximum current I given as
I = Q.ω
ω =angular frequency
By putting the values
ω = 144.33 rad⁻¹
Maximum current
I = 45 x 10⁻³ x 144.33 A
I= 6.49 A
I = 6.5 A
E) 6.5 A
Three 40w lamp for 6 hours
Answer:
Explanation:
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Respuesta:
2 × 10⁴ V
Explicación:
Paso 1: Información provista
- Carga transportada (q): 4 nC
- Trabajo realizado (W): 7 × 10⁻⁵ J
Paso 2: Convertir q a Coulomb
Usaremos el factor de conversión 1 C = 10⁹ nC.
4 nC × 1 C/10⁹ nC = 4 × 10⁻⁹ C
Paso 3: Calcular el potencial eléctrico (V) de la esfera
Usaremos la siguiente fórmula.
V = W/q
V = 7 × 10⁻⁵ J/4 × 10⁻⁹ C = 2 × 10⁴ V
