assuming the reference line to measure the height for gravitational potential energy lying at the equilibrium position
m = mass attached to the spring = 10.00 kg
k = spring constant of the spring = 250 N/m
h = height of the mass above the reference line or equilibrium position = 0.50 m
x = compression of the spring = 0.50 m
v = speed of mass = 2.4 m/s
A = maximum amplitude of the oscillation
v' = speed of mass at the maximum amplitude location = 0 m/s
using conservation of energy between the point where the speed is 2.4 m/s and the highest point at which displacement is maximum from equilibrium
kinetic energy + spring potential energy + gravitational potential energy = kinetic energy at maximum amplitude + spring potential energy at maximum amplitude + gravitational potential energy at maximum amplitude
(0.5) m v² + m g h + (0.5) k x² = (0.5) m v'² + m g A + (0.5) k A²
inserting the values
(0.5) (10) (2.4)² + (10) (9.8) (0.50) + (0.5) (250) (0.50)² = (0.5) (10) (0)² + (10) (9.8) A + (0.5) (250) A²
109.05 = (98) A + (125) A²
A = 0.62 m
The answer is B
because if you compare a regular road and ice... ice is smoother and therefore has less friction
Answer:
0.08 ft/min
Explanation:
To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.
So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:
since the difference between the upper and lower base is the increase in the base and we are only at halft the height.
Now we can calculate the longitudinal section <em>A</em> at that point:
And the raising speed <em>v </em>of the water is given by:
where <em>q</em> is the water flow (1 cubic foot per minute).
As the water russhes toward the shore, it rises because it is pushing against it.<span />