Answer:
a) Length of the arc: [tex]\frac{25\pi }{3}[/tex] [tex]m[/tex] or ≈ 26.18 [tex]m[/tex]
b) Area of the sector: [tex]\frac{125\pi }{6}[/tex] [tex]m^2[/tex] or ≈ 65.45 [tex]m^2[/tex]
c) 16[tex]\pi[/tex] rad or 2880°
Step-by-step explanation:
If a is the length of the arc, r is the radius of the sector and 0 the central angle in radians.
Use the formula:
[set r = 5 m, 0 = [tex]\frac{5\pi }{3}[/tex] (given)]
[tex]a=0[/tex] · [tex]r[/tex]
⇒ [tex]a=\frac{5\pi }{3} .[/tex] [tex]5[/tex]
⇒ [tex]a=\frac{25\pi }{3}[/tex] ≈ [tex]26.18[/tex] [tex]m[/tex]
The arc length of the watered sector is [tex]\frac{25\pi }{3}[/tex] [tex]m[/tex] or ≈ 26.18.
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Recall, the area A of a circle (obviously [tex]0=2\pi[/tex] [tex]rad[/tex]) with a radius [tex]r[/tex] is given by the formula:
[tex]A=\pi r^2=\frac{2\pi }{2} r^2[/tex]
So the area of the circular sector with radius r and central 0 (in radians) will be:
[set r = 5m , 0 = [tex]\frac{5\pi }{3}[/tex] (given)]
[tex]Asector=\frac{0}{2} r^2[/tex]
⇒ [tex]Asector=\frac{5\pi/3 }{2} (5)^2[/tex]
⇒ [tex]Asector[/tex] = [tex]\frac{5\pi }{6}(25)[/tex]
⇒ [tex]Asector[/tex] = [tex]\frac{125\pi }{6}[/tex] ≈ 65.45 [tex]m^2[/tex]
The area of the watered sector is [tex]\frac{125\pi }{6}[/tex] [tex]m^2[/tex] or ≈ 65.45 [tex]m^2[/tex].
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From the given information we know that the sprinkler perform a full revolution (2[tex]\pi[/tex] rad) every 15 sec so we will make the following proportion:
[tex]\frac{15 sec}{ 2 min}[/tex] = [tex]\frac{2\pi }{0}[/tex]
⇒ [tex]\frac{15 sec}{120 sec}[/tex] = [tex]\frac{2\pi }{0}[/tex]
⇒ 0 = [tex]\frac{2\pi . 120 sec}{15 sec}[/tex]
⇒ 0 = [tex]\frac{240\pi }{15}[/tex]
⇒ 0 = [tex]16\pi[/tex]
We can convert the angle of 0 = [tex]16\pi[/tex] rad to degrees as shown:
[tex]0deg[/tex] = 16[tex]\pi[/tex] × [tex]\frac{180°}{\pi }[/tex]
⇒ [tex]0deg[/tex] = 2880°
Finally the sprinkler rotates 16[tex]\pi[/tex] rad or 2880° in 2 minutes.
