Answer:
a.0.120mm
b.1.58s
c.0.6329Hz
Explanation:
a. Given that 0.120mm is displaced from equilibrium, 0.120mm after 0.790s on opposite side:
-The amplitude is the maximum displacement from equilibrium.
-The object goes from x=+A to x=-A and back during one cycle.
#Hence, the amplitude of the motion is 0.120mm
b.Motion from maximum positive displacement to maximum negative displacement takes places during half the period of Simple Harmonic Motion(SHM)
#Hence, the period of the motion is 1.58s
c. Frequency is calculated as one divided by the period of the motion.
From b above we know that the motions period is 1.58s
Therefore:
<em>Frequency=1/period=1/1.58=0.6329Hz</em>
<em>#</em><em>The frequency of the motion is </em><em>0.6329Hz</em>
Answer:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Kinematics of the wheel
Equation of circular motion uniformly accelerated :
ωf = ω₀+ α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
Answer:
521.8 V
Explanation:
The RMS value of the voltage of an AC signal is given by
where
is the peak voltage, which corresponds to the amplitude of the AC waveform
In this problem, we know the RMS voltage
Therefore, we can re-arrange the previous equation to find the peak voltage (the amplitude of the waveform):
Answer:
, downward
Explanation:
There is only one force acting on the ball during its motion: the force of gravity, which is given by
where
m is the mass of the ball
is the acceleration of gravity (downward)
According to Newton's second law,
where F is the net force on the object and a is its acceleration. Rearranging for a,
As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:
Therefore, the ball has a constant acceleration of downward for the entire motion.
