When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
Given:
L = 1 mH =
H
total Resistance, R = 11 
current at t = 0 s,
= 2.8 A
Formula used:

Solution:
Using the given formula:
current after t = 0.5 ms = 
for the inductive circuit:


I =0.011 A
The electrical force acting on a charge q immersed in an electric field is equal to

where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
Answer:
Explanation:
The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.
Change in gravity potential energy = change in spring potential energy
mgh = 1/2kh^2
Assume gravity constant g is 10m/s^2
2.6*10*h = 1/2*200*h^2
100h^2 - 26h = 0
2h(50h - 13) = 0
h = 0 or h = 13/50 = 0.65m
h = 0 is before the spring is stretched
So the maximum distance is 0.65m.
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