#initialisation of variables
H = 1360 #ft
t = 60 #f
a = (10**3)*5.5*(10**-3) #f
q = (1.36*10**3)*5.5*(10**-3) #f
s = (4-1.36)*(10**3)*(3.2*10**-3) #f
#CALCULATIONS
T = t-q-s #F
T1 = T+3*a #F
#RESULTS
print 'the temperature at the mountain top = %.0f F'%(T)
print 'the temperature on the plain beyond the mountain = %.1f F'%(T1)
#initialisation of variables
t = 60 #f
v = 0.52 #in
t1 = 80 #F
p = 40 #percent
v1 = 1.03*0.40 #in
w = 8 #mph
pa = 29.0 #in
p1 = 0.497 #ft
q = 1.32*10**-2 #ft
r = 0.268 #ft
#CALCULATIONS
E = p1*(1-q*pa)*(1+r*w)*(v-.41) #in
#RESULTS
print 'the evaporation for the a day during = %.2e in'%(E)
# note : answer is slightly different because of rounding off error.
#initialisation of variables
t = 47 #f
q = 8000 #ft
a = 100 #ft
d = 0.10 #in
d1 = 7 #degree days
s1 = 14000 #ft
s2 = 7000 #ft
s = 1000 #ft
g = 32 #ft
h = 17.37 #ft
h1 = 1.547 #ft
#CALCULATIONS
T = q+s*(t-g)/3 #ft
T1 = t-3*1 #F
T2 = (T1+g)/2 #F
T3 = d1*d*a #sq mile in
M = h*T3 #mgd
M1 = M*h1 #cfs
#RESULTS
print 'the upper boundary of the melting zone and temperature at the snow line = %.0f F'%(T1)
print 'The average temperature of = %d cfs'%(M1)