A water treatment facility operates by dripping water from a 70 foot long arm whose end is mounted to a central pivot. The water then filters through a layer of charcoal. The arm rotates once every 6 minutes. (Round your answers to two decimal places.) (a) Find the area of charcoal covered with water after 1 minute ft2 (b) Find the area of charcoal covered with water after 1 second ft2 (c) How long would it take to cover 100 square feet of charcoal with water? sec (d) How long would it take to cover 3445 square feet of charcoal with water?

The area of the water treatment facility is the amount of space it covers.

The radius of the water facility is given as:

(a) The area covered in a minute

First, we calculate the area covered in 6 minutes

So, we have:

Divide by 6, to calculate the area covered in 1 minute

Hence, the area covered in 1 minute is 2565.97 ft^2

(b) The area covered in a second

In (a), we have

Divide by 60 to get the area covered in a second

Hence, the area covered in 1 second is 42.77 ft^2

(c) Time to cover 100 ft^2

In (a), we have

and

Cross multiply

Cancel out the common units

Divide both sides by 2565.97

Express as seconds

Hence, it will take 2.34 seconds to cover 100 ft^2

(d) Time to cover 3445 ft^2

In (a), we have

and

Cross multiply

Cancel out the common units

Divide both sides by 2565.97

Hence, it will take 1.34 minutes to cover 3445 ft^2

Read more about areas at:

link

A) the area of charcoal covered with water after 1 minute in ft^(2) = 1309 ft^(2)

B) the area of charcoal covered with water after 1 second (in ft^2) = 21.82 ft^(2)

C) time it would take to cover 100 square feet of charcoal with water (in sec) = 4.58 seconds

D)time it would take to cover 3445 square feet of charcoal with water =

2 min 37.8 seconds

Step-by-step explanation:

A) To find the area of charcoal covered with water after 1 minute in ft^(2); let's determine area that is covered by 1 rotation of the arm.

radius(r) = 50 ft

area(A1) = πr^(2) = 3.14 x 50^(2) = π x 2500 = 7853.98 ft^(2)

Now, from the question, it takes the arm 6 minutes to rotate once.

hence, the area of charcoal covered with water after 1 minute can be found by dividing the total area by 6;

A1/6 = 7853.98/6 = 1309 ft^(2)

B) Now, to find the area of charcoal covered with water after 1 second (in ft^2);

We know that there are 60 seconds in 1 minute, thus we simply divide answer in a above by 60.

= A1/60 = 1309.00/60

A2 = 21.82 ft^(2)

(c) Now, let's find how long it would take to cover 100 square feet of charcoal with water (in sec)

= 100 ft^2 / 7853.98 ft^2 = 0.0127

We know that 6 min =360 sec

Thus; 0.0127 x 360 sec = 4.58 sec

(d) Now, we'll find how long it would take to cover 3445 square feet of charcoal with water?

3445 ft^2 / 7853.98 ft^2 = y / 6 min = y / 360 sec

7853.98y = 360 x 3445

7853.98y = 1,240,200

y = 1,240,200 / 7853.98

y = 157. 91sec = 157.91/60 = 2.63min = 2 min 37.8 seconds

c

step-by-step explanation: