Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;

(a) the force constant of the spring

(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k

Answer:
The electric field is
Explanation:
The force
on a charge
in an electric field
is given by
,
which can be rearranged to give

Now, the force on the electron is
, and its charge is
; therefore,


It is the most massive planet in the solar system.
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Answer:
The average current is 19.567 A
Solution:
As per the question:
Charge, Q = 
Time, t = 
Now,
We know that current is constituted by the rate of transfer of the charge per unit time. Thus we can write:
I =
(1)
Now, the charge that was transferred is 86 % of the original value.
Therefore,
We replace Q by 0.86Q in eqn (1):
I = 