A tank is in the shape of an inverted circular cone with height 500 ft and base radius 20 ft. It is filled with water to a height of 400 ft. Find the work required to empty the tank by pumping all of the water to the top of the tank. (Weight density of water rhog = 62.4 lb/ft3 .)

1.338 * 10 ^ 9 lb-ft

Step-by-step explanation:

We have the following information:

Water density 62.4 lb / ft ^ 3

work equals force per distance, like this:

W = F * d

with the information in the statement we can deduce:

20/500 = r / y

r = y / 25

 now the differential volume would be:

dv = pi * (r ^ 2) * dy

dv = pi * ((y / 25) ^ 2) * dy

dv = pi * (y ^ 2/625) * dy

weight of slice would be:

62.4 * pi * (y ^ 2/625) * dy

= 0.3136 * y ^ 2 * dy

work for the removal of slice:

(500 - y) * 0.3136 * y ^ 2 * dy

Now the total work is the integral of the previous expression that goes from y = 0 to y = 400

Total work = integral (from 0 to 400) {(500 - y) * 0.3136 * y ^ 2 * dy}

Total work = integral (from 0 to 400) {0.3136 * (500 * y ^ 2 - y ^ 3) dy}

We integrate and we have:

Total work = 0.3136 * ((500/3) * y ^ 3 - (y / 4) ^ 4)

Total work = 0.3136 * ((500/3) * 400 ^ 3 - (400/4) ^ 4) - 0.3136 * ((500/3) * 0 ^ 3 - (0/4) ^ 4)

Total work = 0.3136 * ((500/3) * 400 ^ 3 - (400 ^ 4) / 4)

Total work = 1.338 * 10 ^ 9 lb-ft

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

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