
Evaluate These Expressions. The following expressions are somewhat more complex, and use
functions from the math module.
math.sqrt( 40.0/3.0  math.sqrt(12.0) )
6.0/5.0*( (math.sqrt(5)+1) / 2 )**2
math.log( 2198 ) / math.sqrt( 6 )

Run demorandom.py. Run the demorandom.py
script several
times and save the results. Then add the following statement to the
script and run it again several times. What happens when we set an
explicit seed?
#!/usr/bin/env python
import random
random.seed(1)
...everything else the same
Try the following variation, and see what it does.
#!/usr/bin/env python
import random, time
random.seed(time.clock())
...everything else the same

Wind Chill. Wind chill is used by meteorologists to describe the effect of
cold and wind combined.
Given the wind speed in miles per hour,
v
,
and the temperature in °F,
t
, the Wind Chill,
w
, is given by the formula below.
Wind speeds are for 0 to 40 mph, above 40, the difference in
wind speed doesn't have much practical impact on how cold you
feel.
Write a print statement to compute the wind chill felt when it
is 2 °F and the wind is blowing 15 miles per hour.

How Much Does The Atmosphere Weigh? From Slicing Pizzas, Racing Turtles, and Further
Adventures in Applied Mathematics,
[Banks02].
Air Pressure (at sea level)
P_{0}
=
1.01325×10^{5}
newtons/m^{2}. A newton is 1
kg·m/sec^{2} or the force acting on a kg of
mass.
Gravity acceleration (at sea level)
g
=
9.82m/sec^{2}. This converts force from
newtons to kg of mass.
Equation 5.2. Mass of air per square meter (in Kg)
Given the mass of air per square meter, we need to know how many
square meters of surface to apply this mass to.
Radius of Earth
R
=
6.37×10^{6}m.
Equation 5.3. Area of a Sphere
Equation 5.4. Mass of atmosphere (in Kg)
Check: somewhere around
10^{18}kg.

How much does the atmosphere weigh? Part 2. From Slicing Pizzas, Racing Turtles, and Further
Adventures in Applied Mathematics,
[Banks02].
The exercise How Much Does The Atmosphere Weigh?
assumes the earth to be an entirely flat sphere. The averge height of
the land is actually 840m. We can use the ideal gas law to compute the
pressure at this elevation and refine the number a little
further.
Equation 5.5. Pressure at a given elevation
molecular weight of air
m
=
28.96×10^{3} kg/mol.
gas constant
R
= 8.314 joule/°Kmol.
gravity
g
= 9.82
m/sec^{2}.
temperature (°K)
T
= 273°+°C (assume 15°C
for temperature).
elevation
z
= 840 m.
This pressure can be used for the air over land, and the
pressure computed in How Much Does The Atmosphere Weigh? can
be used for the air over the oceans. How much land has this reduced
pressure? Reference material gives the ocean area
(
A_{o}
=
3.61×10^{14} m^{2})
and the land area (
A_{l}
=
1.49×10^{14}
m^{2}).
Equation 5.6. Weight of Atmosphere, adjusted for land elevation