v = 400m/58s = 6.9m/s
v = 400m/60s = 6.7m/s
6.9m/s - 6.7m/s = .2m/s difference
60sec - 58 sec = 2 sec
v =Δx/t
Δx = vt
Δx = (.2m/s)(2sec)
Δx = .4m
therefore, the answer is .4m
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
Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque






Answer:
Tangential velocity = 10.9 m/S
Explanation:
As per the data given in the question,
Force = 20 N
Time = 1.2 S
Length = 16.5 cm
Radius = 33.0 cm
Moment of inertia = 1200 kg.cm^2 = 1200 × 10^(-4) kg.m^2
= 1200 × 10^(-2) m^2
Revolution of the pedal ÷ revolution of wheel = 1
Torque on the pedal = Force × Length
= 20 × 16.5 10^(-2)
= 3.30 N m
So, Angular acceleration = Torque ÷ Moment of inertia
= 3.30 ÷ 12 × 10^(-2)
= 27.50 rad ÷ S^2
Since wheel started rotating from rest, so initial angular velocity = 0 rad/S
Now, Angular velocity = Initial angular velocity + Angular Acceleration × Time
= 0 + 27.50 × 1.2
= 33 rad/S
Hence, Tangential velocity = Angular velocity × Radius
= 33 × 33 × 10^(-2)
= 10.9 m/S