To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is 
Answer:
76.74 Hz
Explanation:
Given:
Wave velocity ( v ) = 330 m / sec
wavelength ( λ ) = 4.3 m
We have to calculate Frequency ( f ):
We know:
v = λ / t [ f = 1 / t ]
v = λ f
= > f = v / λ
Putting values here we get:
= > f = 330 / 4.3 Hz
= > f = 3300 / 43 Hz
= > f = 76.74 Hz
Hence, frequency of sound is 76.74 Hz.
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer:
Magnetic field, B = 0.042 T
Explanation:
It is given that,
Speed of charged particle, 
Angle between velocity and the magnetic field, 
Charge, 
Magnetic force, F = 0.002 N
The magnetic force is given by :
B is the magnetic field
B = 0.042 T
So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.
