Answer:
I think the answer is B. amount of energy present but I'm not 100% sure
Explanation:
I think this is correct, but I am not entirely certain.
Find the force constant of the spring:
F = - KX
(0 - 62.4) = -K(0.172m)
-362.791 = -K
362.791 N/m = K
Find the work done in stretching the spring:
W = (1/2)KX
W = (1/2)(362.791)(0.172m)
W = 31.2 J
Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.

The maximum velocity is

Answer:
Please refer to the figure.
Explanation:
The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.
So,

This enclosed current is now to be used in Ampere’s Law.

Here,
represents the circular path of radius r. So we can replace the integral with the circumference of the path,
.
As a result, the magnetic field is

Answer:
The relationship between the wave's amplitude and frequency is such that it is inversely proportional to the frequency. The amplitude decreases as the frequency increases. The amplitude increases as the frequency decreases. The higher the energy of a wave, the higher the amplitude. The lower the energy, the lower the amplitude. Energy has no effect on wavelength, speed, or frequency, only the amplitude.
Explanation: