## Transcribed Text

1. The tide in 4 kilometer deep water has twice the celerity of the tide in 1 km of
water. From the equations for wave celerity, show if this is true or false.
2. How deep is the Bay of Fundy, if it is 200 kilometers long and acts as a quarter
wavelength resonator, resonant with a 12 hour 25 minute tide?
The relevant equations are celerity C= 2/T = (g H)1/2 ==1/4, so T=4L (g H>-1/22
solve for H.
L
sea
s.riace
H
Depth
of
Bay of
Fundy
Bay of Fundy length
3. A tsunami has a wavelength ~ 100 kilometers. What is the celerity in 4
kilometers of water? What is the celerity in 10 meters of water?
4. What is the volume transport of the Gulf Stream in Sverdrup units if the Gulf
Stream is 100 km wide and 1.5 kilometers deep, at a velocity of 1.5 meters per
second?
5. In a developed sea of deep water waves with.
Wind speed:
20 km/hr
40 km/hr
80 km/hr
Average height:
0.33 m
1.8 m
10.3 m
Wavelength
10.6m
22.2m
158.6m
Compute the celerity of each average wave.
6. Compute which waves are breaking waves in # 5.
7. You are fishing in 4 meters of water. There is an ocean swell passing
regularly. The period is about 10 seconds.
a) What is the celerity of the swell in the open-ocean?
b) What is the celerity of the wave passing under your boat?
8. What is the volume transport in Sverdrup units of the equatorial undercurrent?
Assume it to be a jet lying between 50 and 250 meters depth, width of 100 km
and length of 12000 km, with celerity of 2.5 m/sec? How does this compare with
the Gulf stream above?
9. What is the value of the Coriolis parameter at the pole, using the equations
given in slides 21 and 27, lecture on advective processes?
10. The ocean has a volume of 1x109 km³, as you computed in the previous
problem set. If ice loss for Greenland and Antarctica is 500 km³ per annum, how
much does the ocean level rise from that source?

This material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.

4) One Sverdrup is equal to one million cubic meters per second.

(100000) (1500) (1.5) = 225000000 cubic meters per second.

This is equivalent to 225 Sverdrups.

5) The celerities are:

(9.8 * 10.6 / (2pi))^(1/2) = 4.07

(9.8 * 22.2 / (2pi))^(1/2) = 5.88

(9.8 * 158.6 / (2pi))^(1/2) = 15.73

6)

0.33 / 10.6 = 0.031

not breaking

1.8 / 22.2 = 0.081

breaking

10.3 / 158.6 = 0.0649

not breaking...