Respuesta :
Answer:
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
The area of the water treatment facility is the amount of space it covers.
- The area covered in 1 minute is 2565.97 ft^2
- The area covered in 1 second is 42.77 ft^2
- It will take 2.34 seconds to cover 100 ft^2
- It will take 1.34 minutes to cover 3445 ft^2
The radius of the water facility is given as:
[tex]\mathbf{r = 70ft}[/tex]
(a) The area covered in a minute
First, we calculate the area covered in 6 minutes
[tex]\mathbf{Area =\pi r^2}[/tex]
So, we have:
[tex]\mathbf{Area =3.142 \times 70^2}[/tex]
[tex]\mathbf{Area =15395.8ft^2}[/tex]
Divide by 6, to calculate the area covered in 1 minute
[tex]\mathbf{1\ min= \frac{15395.8ft^2}6}[/tex]
[tex]\mathbf{1\ min= 2565.97\ ft^2}[/tex]
Hence, the area covered in 1 minute is 2565.97 ft^2
(b) The area covered in a second
In (a), we have
[tex]\mathbf{1\ min= 2565.97\ ft^2}[/tex]
Divide by 60 to get the area covered in a second
[tex]\mathbf{1\ sec= \frac{2565.97\ ft^2}{60}}[/tex]
[tex]\mathbf{1\ sec= 42.77\ ft^2}[/tex]
Hence, the area covered in 1 second is 42.77 ft^2
(c) Time to cover 100 ft^2
In (a), we have
[tex]\mathbf{1\ min= 2565.97\ ft^2}[/tex]
and
[tex]\mathbf{Time =100ft^2}[/tex]
Cross multiply
[tex]\mathbf{Time \times 2565.97ft^2=100ft^2 \times 1\ min}[/tex]
Cancel out the common units
[tex]\mathbf{Time \times 2565.97=100 \times 1\ min}[/tex]
Divide both sides by 2565.97
[tex]\mathbf{Time =0.03897\ min}[/tex]
Express as seconds
[tex]\mathbf{Time =0.03897\ \times 60\ seconds}[/tex]
[tex]\mathbf{Time =2.34\ seconds}[/tex]
Hence, it will take 2.34 seconds to cover 100 ft^2
(d) Time to cover 3445 ft^2
In (a), we have
[tex]\mathbf{1\ min= 2565.97\ ft^2}[/tex]
and
[tex]\mathbf{Time =3445ft^2}[/tex]
Cross multiply
[tex]\mathbf{Time \times 2565.97ft^2=3445ft^2 \times 1\ min}[/tex]
Cancel out the common units
[tex]\mathbf{Time \times 2565.97=3445 \times 1\ min}[/tex]
Divide both sides by 2565.97
[tex]\mathbf{Time =1.34\ min}[/tex]
Hence, it will take 1.34 minutes to cover 3445 ft^2
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