A source injects an electron of speed v = 2.9 × 107 m/s into a uniform magnetic field of magnitude B = 1.7 × 10-3 T. The velocit
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The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:
x = 0.135 cm
Given parameters
- The mass m = 220 g = 0.220 kg
- The spring cosntnate3 k = 7.0 N / m
- Initial displacement A = 5.2 cm = 5.2 10-2 m
To find
- The position where the kinetic and potential energy are equal
A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.
x = A cos wt + fi
w² =
Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.
The speed is defined by the variation of the position with respect to time.
v =
let's evaluate
v = - A w sin (wt + Ф)
Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.
0 = - A w sin Ф
For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero
x = A cos wt
Let's find the point where the kinetic and potential energy are equal.
K = U
½ m v² = m g x
we substitute
½ A² w² sin² wt = g A cos wt
sin² wt = cos wt
let's calculate
w =
w = 5.64 rad / s
sin² 5.64t = 2 9.8 / 0.052 cos 5.64t
sin² 5.64t = 376.92 cos 5.64 t
1 - cos² 5.64t = 376.92 cos 5.64t
cos² 5.64t -376.92 cos564t -1 = 0
we make the change of variable
x = cos 5.64t
x²- 376.92 x - 1 = 0
x = 0.026
cos 5.64t = 0.026
Let's find the displacement for this time
x = 5.2 10-2 0.026
x = 1.35 10-3 m
In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:
x = 0.135 cm
Learn more here: brainly.com/question/15707891
Explanation:
Given that,
Resistance R = 75.0 ohms
Inductance L = 55.0 mH
Capacitance
Voltage V = 12.0 V
Frequency f = 60.0 Hz
We need to calculate the angular frequency
Using formula of angular frequency
Put the value into the formula
(a). We need to calculate the value of
Using formula of
Put the value into the formula
(b). We need to calculate the value of
Using formula of
(c). We need to calculate the value of Z
Using formula of impedance
Put the value into the formula
(d). We need to calculate the rms current
Firstly we need to calculate the current
Using formula of current
Put the value into the formula
Using formula of rms current
(e). We need to calculate the rms voltage across the resistor
Using formula of rms voltage
(f). We need to calculate the rms voltage across the inductor
Using formula of rms voltage
(g). We need to calculate the rms voltage across the capacitor
Using formula of rms voltage
(h). We need to calculate the dissipated power by the circuit
Using formula of dissipated power
Put the value into the formula
Hence, This is the required solution.