the axis acts against and it would be a contact force
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
At the bottom of the tank :
P = ρgH
P = (1000 kg/m³)(10 m/s²)(1 m)
P = 10000 N/m²
F = P • A
F = (10000 N/m²)(1 m²)
F = 10000 N
At the side of the tank :
Pav = ½ρgH
Pav = ½(1000 kg/m³)(10 m/s²)(1 m)
Pav = 5000 N/m²
F = P • A
F = (5000 N/m²)(1 m²)
F = 5000 N
I believe it’s called Alluvium! It’s where the river mouth is build up of gravel,sand,silt, and clay!!