Physical Oceanography / Tide Problem

[PhysO]

Homework Statement

Two ports along a coastline are 300 km apart. One port has high tides 30 minutes later than the other. How deep is the effective depth for the tidal propagation?

Homework Equations

I am thinking I need the following equation: V = square root (g*H). V = velocity. g = 9.8 m/s^2. H = the depth of tidal propagation.

The Attempt at a Solution

I calculated the velocity (300 km/ 30 min = 167 m / s). Then I entered that into the equation, V = square root (g*H). I got H = 2845.8 m. I am wondering if this is the right approach for this problem. It seems too simple.

 

[nate808]

Your approach is correct, but there are a few things you can clarify to make your solution more complete.

First, it would be helpful to specify the units for your velocity calculation. In this case, the units should be in meters per second (m/s).

Second, you should also specify the units for your depth calculation. In this case, the units should be in meters (m).

Additionally, it would be helpful to explain why you used the given equation, V = square root (g*H), and how it relates to the problem. This equation is known as the shallow water wave speed equation, and it is commonly used to calculate the speed of waves in shallow bodies of water. In this case, the wave speed (V) is equal to the square root of the product of the acceleration due to gravity (g) and the depth of the water (H). This equation is relevant to the problem because it allows us to calculate the depth of the tidal propagation based on the known wave speed and acceleration due to gravity.

Finally, to make your solution more clear and concise, you can write your final answer as follows:

The effective depth for the tidal propagation is 2845.8 meters. This was calculated using the shallow water wave speed equation, V = square root (g*H), with a wave speed of 167 m/s and an acceleration due to gravity of 9.8 m/s^2.