A mass attached to a 66.4 cm long string starts from rest and is rotated 36.6 rev in1 min before reaching a final angular speed.
1 answer:
Answer:
0.128 rad/s², 7.66 rad/s
Explanation:
length, l = 66.4 cm
initial angular velocity, ωo = 0 rad/s
Let ω represent the final angular velocity.
Let α denote the angular acceleration.
number of revolutions, n = 36.6
time taken, t = 1 min = 60 seconds
Angle rotated, θ = 2πn = 2 x 3.14 x 36.6 = 229.85 rad
Apply the second equation of motion for angular dynamics
229.85 = 0 + 0.5 x α x 60 x 60
α = 0.128 rad/s²
Utilize the first equation of motion
ω = ωo + αt
ω = 0 + 0.128 x 60
ω = 7.66 rad/s
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Complete Question
An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.
At time 5 seconds, I = 1.2 A.
We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.
Answer:
The charge is
Explanation:
The question indicates that
The wire’s diameter is
The radius of the wire is
Aluminum's resistivity is
The electric field variation is described as
The charge is effectively given by the equation
Where A is the area expressed as
Thus,
Therefore
By substituting values
The question states that t = 5 seconds
Answer:
La magnitud del EMF es 0.00055 volts
Explanation:
El EMF inducido es proporcional al cambio en el flujo magnético según la ley de Faraday:
Como en nuestro caso hay solo un lazo de alambre, entonces N=1 y obtenemos:
Necesitamos expresar el flujo magnético dada la geometría del problema;
donde A es el área de la bobina que permanece constante con el tiempo, y B es el campo magnético que cambia con el tiempo. Por lo tanto, la ecuación para el EMF se convierte en: