Activity Answers
Broadcast 5


Weekly Weather Log
Each week please complete the following information for both our location here off the coast of Australia and the location at your school. Each week students should complete the weekly log with the weather information at their school. Mrs. Linsley will send the ship weather report by email following each broadcast. Students can record this information at that time. Also, the front page of the web site will always display the time and date in Australia.

  My School Ship in Australia

Temperature

*
28°C

Day and Time

*
February 14, 2001 @ 10:00am
Precipitation
[indicate snow, sleet or rain]
*
None
Skies or Cloud Bank
*
5/8
Season
*
Summer
Wind Speed and Direction
*
14 Knots, East


Formula for Converting Celsius to Fahrenheit:
F= (1.8 X C) +32
C=F-32/1.8

1 Knot = 1 nautical mile per hour (1 nautical mile = 6076 feet or 1852 meters)

Cloud Cover is recorded by a code which lists the coverage in eights:
Code:Eights
0:0
1:1/8 or less
2:2/8
3:3/8
4:4/8
5:5/8
6:6/8
7:7/8
8:8/8 Overcast
9: obscured or cloud amount cannot be determined

Questions/Activities
[Complete the questions below prior to the broadcasts, if at all possible]

Given the equation: velocity = distance/time

If we know that it takes 0.7 seconds for the sound wave to travel the seafloor and back, and that the velocity of sound in the ocean is 1.4 kilometers/second, what is the depth of the water?
To find the answer, first solve the equation for distance:
Distance = velocity * time
Find the time it took the wave to travel to the seafloor from the ship:
Time to Seafloor = Total time/2
Time to Seafloor = 0.7 sec/2 = 0.35 sec
Substitute the information: Distance = 1.4 km/sec * 0.35 sec
Solve: Distance = 0.49 km
To get the answer in meters (if desired):
Distance = 0.49 km * 1000 m/km = 490 meters To get the answer in feet (if desired): Distance = 490 m * 3.25 ft/m = 1592.5 feet

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