Answer
given,
Two solenoids A and B
Number of turn
Na = 430 turns Nb = 610 turns
Current = 2.80 A
Average flux through A = 300 μWb
Average of flux through B = 90.0 μ
Wb
a) 


b) inductance of A



c) magnitude of the emf




I believe the answer is B. that's where the asetroid belt is.
Answer:
I don't know about this, this is which class question
Answer:

Explanation:
We first identify the elements of this simple harmonic motion:
The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.
The angular frequency ω can be calculated with the formula:

Where k is the spring constant and m is the mass of the particle.
Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

Finally, the equation of the motion of the system is:
or
