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the diameter of a circular swimming pool is 18 meters what is the problem


Sagot :

Step-by-step explanation:

Consider a thin horizontal circular cross-sectional slice of the water which is cylindrical in shape having thickness or height

Δ

y

,

with radius 9 m at y m above the bottom of the pool. Then the volume of this cylindrical slice will be given by

V

o

l

u

m

e

=

π

×

r

a

d

i

u

s

2

×

h

e

i

g

h

t

=

π

(

9

2

)

Δ

y

=

81

π

Δ

y

.

Then the force exerted by the slice will be given by

F

o

r

c

e

=

M

a

s

s

×

A

c

c

e

l

e

r

a

t

i

o

n

=

V

o

l

u

m

e

×

D

e

n

s

i

t

y

×

A

c

c

e

l

e

r

a

t

i

o

n

(

g

)

=

9.8

(

1000

)

81

π

Δ

y

.

So the work done in pumping this slice of water out of the top of the pool, through a distance (4 - y) meters, will be given by

W

o

r

k

=

F

o

r

c

e

×

D

i

s

t

a

n

c

e

=

M

a

s

s

×

A

c

c

e

l

e

r

a

t

i

o

n

=

V

o

l

u

m

e

×

D

e

n

s

i

t

y

×

A

c

c

e

l

e

r

a

t

i

o

n

(

g

)

=

9.8

(

1000

)

81

π

(

4

y

)

Δ

y

(

1

)

So, using (1), the work done in pumping ALL the water, at height 2.5 m, over the top of the pool will be given by the definite integral:

T

o

t

a

l

W

o

r

k

=

2.5

0

9.8

(

1000

)

81

π

(

4

y

)

d

y

=

9.8

(

1000

)

81

π

2.5

0

(

4

y

)

d

y

=

9.8

(

1000

)

81

π

[

4

y

y

2

2

]

2.5

0

=

17144849.21.

The work done in pumping the water out of the side of the pool is 17144849.21 Joules (J).

(b) When pumping the water out of an outlet 1 meter above the top of the pool, the height to which the water needs to be pumped to, increases to 5 m. Hence (1) above will change to

W

o

r

k

=

F

o

r

c

e

×

D

i

s

t

a

n

c

e

=

M

a

s

s

×

A

c

c

e

l

e

r

a

t

i

o

n

=

V

o

l

u

m

e

×

D

e

n

s

i

t

y

×

A

c

c

e

l

e

r

a

t

i

o

n

(

g

)

=

9.8

(

1000

)

81

π

(

5

y

)

Δ

y

(

2

)

So, using (2), the work done in pumping ALL the water, at height 2.5 m, through the outlet will be given by the definite integral:

T

o

t

a

l

W

o

r

k

=

2.5

0

9.8

(

1000

)

81

π

(

5

y

)

d

y

=

9.8

(

1000

)

81

π

2.5

0

(

5

y

)

d

y

=

9.8

(

1000

)

81

π

[

5

y

y

2

2

]

2.5

0

=

23379339.83.

The total work in this case will be 23379339.83 Joules (J).