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
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N
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Answer:
1) No, the car does not travel at constant speed.
2) V = 9 ft/s
3) No, the car does not travel at constant speed.
4) V = 5.9 ft/s
Explanation:
In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:
V(t) = 2*t + 2 Since the speed depends on time, the speed is not constant at any time.
For the average speed we evaluate the formula for t=2 and t=5:
d(2) = 8 ft and d(5) = 35 ft

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:
d(1.8) = 6.84 ft and d(2.1) = 8.61 ft

dissipation is the answer ;(