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:
As described, The mantle causes the tectonic plates to move.
Answer:
A) 0 miles
Explanation:
The displacement is the distance between the starting point and the end point of the route.
In this case, even if Matino takes a whole tour around the city, since<u> it ends in the same place where it started</u>, the difference between the starting and finishing point is zero, so its total displacement is zero.
Care must be taken to distinguish the terms of displacement and distance traveled, beacause they are not the same, since in this case the distance traveled would be 3.15miles, but the displacement is zero, because it ends at the point where it started.
Answer:
The strength of magnetic field is 0.2 Tesla.
Explanation:
Data from the question is
Length (L) of wire ; L=0.10 m
Current in wire ; I= 2.0 A
Force on wire ; F = 0.04 N
Angle = Right angle So, 

Now ,
We have to find the magnetic Field strength (B)
For this formula for Force on wire in magnetic field is

Further modified as

Now insert values in the formula


So, the strength of magnetic field is 0.2 Tesla.