A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
1 answer:
Answer:
The value is
Explanation:
From the question we are told that
The mass of the block is
The force constant of the spring is
The amplitude is
The time consider is
Generally the angular velocity of this block is mathematically represented as
=>
=>
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
=>
=>
You might be interested in
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + ( -
)² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL
Answer:
Explanation:
You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:
where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.
by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:
(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.
By doing the same for the electron you obtain:
we can equals these expressions for both proton and electron, because the forces qE are the same:
Answer:
The magnitude of the electric field at a point equidistant from the lines is
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
