Crests of an ocean wave pass a pier every 10.0 s. if the waves are moving at 5.6 m/s, what is the wavelength of the ocean waves?
1 answer:
Every 10.0 seconds, a crest of the wave passes the pier. This means that the period of the wave is exactly 10.0 s:
which means that the frequency of the wave is
The wavelength of a wave is related to its frequency by the relationship
where v is the speed of the wave.
In this problem, v=5.6 m/s; if we use the previous formula, we find the wavelength of the wave:
which means that the frequency of the wave is
The wavelength of a wave is related to its frequency by the relationship
where v is the speed of the wave.
In this problem, v=5.6 m/s; if we use the previous formula, we find the wavelength of the wave:
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<h2>Answer:</h2>
(a) 10N
<h2>Explanation:</h2>The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
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Answer
given,
diameter,d₁ = 7.5 cm
d₂ = 4.5 cm
P₁ = 32 kPa
P₂ = 25 kPa
Assuming, we have calculation of flow in the pipe
using continuity equation
A₁ v₁ = A₂ v₂
π r₁² v₁ = π r₂² v₂
Applying Bernoulli's equation
v₂ = 4.01 m/s
fluid flow rate
Q = A₂ V₂
Q = π (0.0225)² x 4.01
Q = 6.38 x 10⁻³ m³/s
flow in the pipe is equal to 6.38 x 10⁻³ m³/s