a drainage basin bordering the sea has an area of 7500 km2 . the average precipitation for this drainage basin is 900 mm year-1 . the average surface water flow at the outlet of the drainage basin equals 22.5 × 108 m3 year-1 . the average groundwater flow to the sea is 100 mm year-1 . the averages are determined for 30 hydrological years.

a. draw up the water balance.

b. determine the average actual evaporation in mm year-1 and m3 year-1

Answer with Step-by-step explanation:

The basic water balance dictates that "Amount of water flowing into the basin should be equal to the amount of water flowing out of it".

Mathematically

where

P = precipitation

I = Infiltration losses in the basin

E = Evapotranspiration

Q = is outflow from the basin

Applying the given values we get

Part b)

From the above equation we get

The evaporation in cubic meters per year is

215.92 dollars is the answer

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841