What is the vertical distance between a trough and a crest?
2 answers:
Answer:
d. Wave Height
Explanation:
Imagine my crude drawing of a sea wave is actually passable haha, so, the top of the wave is called the crest, and the lowest part of the wave, under the calm sea level, is the Trough, the vertical distance between those two points is the wave height.
As extra credit ;) the distance between two consecutive crests or two consecutive troughs is the wavelength, and this is exactly the same in radio waves!
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Answer:
b > 66.41 kg/s
Explanation:
The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f
F + f = ma
-kx - bv = ma
-kx -bdx/dt = md²x/dt².
Re-arranging the equation, we have
So, md²x/dt² + bdx/dt + kx = 0
Dividing through by m, we have
d²x/dt² + (b/m)dx/dt + (k/m)x = 0
This is a second-order differential equation. The characteristic equation is thus,
D² + (b/m)D + (k/m) = 0
Using the quadratic formula, we find D.
For an overdamped system,
Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.
Substituting k = mg/x into the inequality for b, we have
b > 2√{(mg/x₀)m}
b > 2√{(m²g/x₀)}
b > 2m√{g/x₀)}
b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}
b > 1.5 kg√{19.6/s²)}
b > 1.5 kg × 4.427/s
b > 66.41 kg/s
Answer:
Combined speed, V = 2 m/s
Explanation:
Given that,
Mass of object 1, m = 5 kg
Speed of object 1, u = 4 m/s (due right)
Mass of object 2, m' = 2 kg
Speed of object 2, u' = -3 m/s (due left)
<em>Let V is the speed of the objects after the collision. Here, both objects stick to each other. Therefore, the momentum remains conserved. So,</em>
So, the combined speed of the masses will be 2 m/s.