Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo
si se sabe que w=2pi f
1 answer:
Responder:
A) ω = 565.56 rad / seg
B) f = 90Hz
C) 0.011111s
Explicación:
Dado que:
Velocidad = 5400 rpm (revolución por minuto)
La velocidad angular (ω) = 2πf
Donde f = frecuencia
ω = 5400 rev / minuto
1 minuto = 60 segundos
2πrad = I revolución
Por lo tanto,
ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)
ω = (5400 * 2πrad) / 60 s
ω = 10800πrad / 60 s
ω = 180πrad / seg
ω = 565.56 rad / seg
SI)
Dado que :
ω = 2πf
donde f = frecuencia, ω = velocidad angular en rad / s
f = ω / 2π
f = 565.56 / 2π
f = 90.011669
f = 90 Hz
C) Periodo (T)
Recordar T = 1 / f
Por lo tanto,
T = 1/90
T = 0.0111111s
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