On a sunny day in an estuary, the surface irradiance is 1.6*10^17 photons cm^-2 s^-1. What is k if the irradiance is 7.3*10^13 photons cm^-1 s^-1 at 16 m depth?
At a spot in the ocean, the surface irradiance is 7.8*10^16 photons and k=.14 m^-1. At what depth does the euphotic zone end?
Your airplane leaves 20 degrees N(linear velocity= 845 nm/h) heading to a point due north at 25 degrees north (linear velocity = 815 nm/h). You set the autopilot dor due north and go to sleep. If your flight time is 30 minutes, how far off and in what direction are you if you take no corrective action?
What is the instantaneous acceleration (both magnitude and direction) due to the coriolis force on an object moving at 160 m s^-1 to the west located at 17 degrees south, 86 degrees west?
In a fit of pique, the Canadians fire a cannonball due south from vancouver (lat 49.2 degrees N) at a randomly picked city (39.4 degrees N). Assuming the cannonball takes 48 minutes to get there, by how far and in what direction will the cannonball miss?
If you had posted these questions separately, an answer might have been obtained sooner. I can only help with some of your questions. Oceanography is not my field.
(a) For the first one, solve this "Beer's law" equation for k:
exp(-kL) = 7.3*10^13/1.6*10^17 = 4.56*10^-4
kL = 7.7
k = 7.7/16 = 0.48 m^-1
This assumes the sun is overhead. It usually isn't.
(b) Use the same equation, but the lower value of k. Use the definition of the the "euphotic zone" (whatever that is) to solve for L
At extreme depths, Beer's law may not apply because scattering becomes important. You also need to know the elevation angle of the sun.
I can only guess that we are talking about the decay constant in I = Io e^-kx
Where for example :
Io = 1.6 * 10^17 *10^4 photons/m^2s
and
i at 16 m = 7.3 * 10^13 * 10^4 = 1.6 * 10^17 * 10^4 * e^-k(16)
4.56 *10^-4 = e^-16k
ln 4.56 -4 ln10 = -16 k
1.517 - 9.210 = -7.693 =16 k
k = 0.481/m
For the second part I do not know how dark it is when we say the euphotic zone ends. I am sure the irradiance for that is defined in whatever text the student is using.
To solve these questions, we will need to apply some concepts and formulas. Let's go through each question and explain how to get the answer.
1) On a sunny day in an estuary, the surface irradiance is 1.6*10^17 photons cm^-2 s^-1. We are given the irradiance at a depth of 16 m, which is 7.3*10^13 photons cm^-1 s^-1. We need to find the attenuation coefficient, k.
The formula to calculate the attenuation coefficient is:
k = -ln(I/I0) / d
Where:
I is the irradiance at a specific depth
I0 is the surface irradiance
d is the depth
Using the given values, we can substitute them into the formula to solve for k.
2) At a spot in the ocean, the surface irradiance is 7.8*10^16 photons. We are given k = 0.14 m^-1. We need to find the depth at which the euphotic zone ends.
The euphotic zone is the depth at which the light intensity is reduced to 1% of its surface value. To calculate it, we can use the formula:
Depth = log(0.01) / -k
Substituting the given value of k into the formula will give us the depth at which the euphotic zone ends.
3) Your airplane leaves 20 degrees N(linear velocity = 845 nm/h) heading to a point due north at 25 degrees north (linear velocity = 815 nm/h). We need to find how far off and in what direction you are if you take no corrective action.
To solve this question, we need to calculate the distance and direction between the starting and ending points. We can use the concept of vector addition and trigonometry to find the answers.
First, we need to calculate the time of flight for the journey, which is given as 30 minutes. Next, we can use the formulas of velocity and time to find the distance traveled by each segment. After that, we can use these distances, along with the angles provided, to calculate the overall displacement or how far off you are and in what direction.
4) We are given the velocity of the object moving at 160 m s^-1 to the west located at 17 degrees south, 86 degrees west. We need to find the instantaneous acceleration (both magnitude and direction) due to the Coriolis force.
The Coriolis force is given by the formula:
F = 2 * m * v * sin(Φ)
Where:
F is the Coriolis force
m is the mass of the object
v is the velocity of the object
Φ is the latitude
We can substitute the given values into the formula to find the magnitude of the Coriolis force. The direction of the force will be perpendicular to the velocity vector and the axis of rotation (in this case, to the right of the direction of motion due to being in the southern hemisphere).
5) In this question, the Canadians fire a cannonball due south from Vancouver (lat 49.2 degrees N) at a randomly picked city (lat 39.4 degrees N). We are given the flight time of 48 minutes. We need to find by how far and in what direction will the cannonball miss.
To calculate the distance and direction of the miss, we can use the concept of projectile motion. We need to calculate the initial velocity of the cannonball and its range or horizontal distance traveled in the given time. Once we have these values, we can calculate the difference between the desired target distance and the actual range of the cannonball to find the miss distance. The direction of the miss will depend on the latitude difference between Vancouver and the target city.
By applying these concepts and formulas, we can solve each question step by step and find the desired answers.