<h3><u>Given Information :</u></h3>
- Radius of the turbine = 32 m
- π = 3
<h3 /><h3><u>To calculate :</u></h3>
- Area covered by blade when the turbine makes 1 revolution.
<h3><u>Calculation :</u></h3>
We know that,
![\bigstar \: \boxed{\sf {Area_{(Circle)} = \pi r^2}} \\](https://tex.z-dn.net/?f=%5Cbigstar%20%5C%3A%20%5Cboxed%7B%5Csf%20%7BArea_%7B%28Circle%29%7D%20%3D%20%5Cpi%20r%5E2%7D%7D%20%5C%5C%20)
![\sf \longmapsto {Area = \{3 \times (32)^2 \} \: m^2 }](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%7BArea%20%3D%20%5C%7B3%20%5Ctimes%20%2832%29%5E2%20%5C%7D%20%5C%3A%20m%5E2%20%7D%20)
![\sf \longmapsto {Area = \{3 \times 1024 \} \: m^2 }](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%7BArea%20%3D%20%5C%7B3%20%5Ctimes%201024%20%5C%7D%20%5C%3A%20m%5E2%20%7D%20)
![\longmapsto \sf \red {Area = 3072 \: m^2 }](https://tex.z-dn.net/?f=%20%5Clongmapsto%20%5Csf%20%5Cred%20%20%7BArea%20%3D%203072%20%5C%3A%20m%5E2%20%7D%20)
Therefore,
- Correct option is <u>option 4th.</u>
Answer:
1. d<1 d less than 1
2.q<3 (q less than 3)
3.c>5 (c greater than)
4.d<d+3 (d less than d plus 3)
5.c<d+3 (c less than d plus 3)
6.q<q+2 (q less than q plus q)
Step-by-step explanation:
Answer: There will enough to paint the outside of a typical spherical water tower.
Step-by-step explanation:
1. Solve for the radius r from the formula for calculate the volume of a sphere. as following:
![V=\frac{4}{3}r^{3}\pi\\\frac{3V}{4\pi}=r^{3}\\r=\sqrt[3]{\frac{3V}{4\pi}}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7Dr%5E%7B3%7D%5Cpi%5C%5C%5Cfrac%7B3V%7D%7B4%5Cpi%7D%3Dr%5E%7B3%7D%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%7D%7D)
2. Substitute values:
![r=\sqrt[3]{\frac{3(66,840.28ft^{3})}{4\pi}}=25.17ft](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3%2866%2C840.28ft%5E%7B3%7D%29%7D%7B4%5Cpi%7D%7D%3D25.17ft)
3. Substitute the value of the radius into the equation fo calculate the surface area of a sphere, then you obtain that the surface area of a typical spherical water tower is:
![SA=4r^{2}\pi\\SA=4(25.17ft)^{2} \pi\\SA=7,964.95ft^{2}](https://tex.z-dn.net/?f=SA%3D4r%5E%7B2%7D%5Cpi%5C%5CSA%3D4%2825.17ft%29%5E%7B2%7D%20%5Cpi%5C%5CSA%3D7%2C964.95ft%5E%7B2%7D)
3. If a city has 25 gallons of paint available and one gallon of paint covers 400 square feet of surface area, you must multiply 25 by 400 square feet to know if there will be enough to paint the outside of a typical spherical water tower.
![25*400ft^{2}=10,000ft^{2}](https://tex.z-dn.net/?f=25%2A400ft%5E%7B2%7D%3D10%2C000ft%5E%7B2%7D)
As you can see, there will enough to paint the outside of a typical spherical water tower.
Answer:
3
Step-by-step explanation:
A ball is 10 meters. Since 100 centimeters are in a meter, each ball has 100*10 = 1000 centimeters.
The teacher needs 70 strings each 40 centimeters. The total amount they need is 70*40 = 2800 centimeters.
Since a ball has 1000 centimeters, 2800 centimeters will need 3 balls for a total of 3000 centimeters.