Answer:
Magnetic field in Tesla: 1.65*10^−10 T, 0, 3.09*10^−10 T
Explanation:
Given data:
i = 4.8 A ,
dBx = 1.65*10^-10 T
dBy = 0 T
dBz = 3.09 * 10^-10 T
Attached below is the detailed solution
When you stretch a spring 20 cm past its natural length, it exerts a force of 8
2 answers:
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a)
The equation which gives the position of a simple harmonic oscillator is:
where
A is the amplitude
is the angular frequency, with k being the spring constant and m the mass
t is the time
Let's start by calculating the angular frequency:
The amplitude, A, can be found from the maximum velocity of the spring:
So, the equation of motion is
b) t=0.10 s, t=0.52 s
The potential energy is given by:
While the kinetic energy is given by:
The velocity as a function of time t is:
The problem asks as the time t at which U=3K, so we have:
However, , so we have
with two solutions:
c) 3 seconds.
When x=0, the equation of motion is:
so, t=0.
When x=1.00 m, the equation of motion is:
So, the time needed is 3 seconds.
d) 0.097 m
The period of the oscillator in this problem is:
The period of a pendulum is:
where L is the length of the pendulum. By using T=0.628 s, we find