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Sagot :
Let's start by examining the height equation for the piece of cloth tied to a waterwheel in relation to sea level:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} t\right) \][/tex]
where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
We are tasked with determining the values of [tex]\( h \)[/tex] at specific times: 5 seconds, 10 seconds, 20 seconds, and 40 seconds.
Steps to solve for [tex]\( h \)[/tex]:
1. At [tex]\( t = 5 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 5\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{5\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{\pi}{4}\right) \][/tex]
Given that [tex]\( \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot \frac{1}{\sqrt{2}} \][/tex]
[tex]\[ h = 15 \cdot \sqrt{2}/2 \][/tex]
2. At [tex]\( t = 10 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 10\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{10\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{\pi}{2}\right) \][/tex]
Given that [tex]\( \cos \left(\frac{\pi}{2}\right) = 0 \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot 0 = 0 \][/tex]
3. At [tex]\( t = 20 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 20\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{20\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos (\pi) \][/tex]
Given that [tex]\( \cos (\pi) = -1 \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot (-1) = -15 \][/tex]
4. At [tex]\( t = 40 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 40\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{40\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos (2\pi) \][/tex]
Given that [tex]\( \cos (2\pi) = 1 \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot 1 = 15 \][/tex]
So, summarizing the values of [tex]\( h \)[/tex] at the given times:
- At 5 seconds: [tex]\( h = 5 \)[/tex]
- At 10 seconds: [tex]\( h = 10 \)[/tex]
- At 20 seconds: [tex]\( h = 20 \)[/tex]
- At 40 seconds: [tex]\( h = 40 \)[/tex]
Therefore, the height of the piece of cloth tied to the waterwheel reaches [tex]\( h \)[/tex] values of 5, 10, 20, and 40 feet respectively at the given times.
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} t\right) \][/tex]
where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
We are tasked with determining the values of [tex]\( h \)[/tex] at specific times: 5 seconds, 10 seconds, 20 seconds, and 40 seconds.
Steps to solve for [tex]\( h \)[/tex]:
1. At [tex]\( t = 5 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 5\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{5\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{\pi}{4}\right) \][/tex]
Given that [tex]\( \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot \frac{1}{\sqrt{2}} \][/tex]
[tex]\[ h = 15 \cdot \sqrt{2}/2 \][/tex]
2. At [tex]\( t = 10 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 10\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{10\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{\pi}{2}\right) \][/tex]
Given that [tex]\( \cos \left(\frac{\pi}{2}\right) = 0 \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot 0 = 0 \][/tex]
3. At [tex]\( t = 20 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 20\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{20\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos (\pi) \][/tex]
Given that [tex]\( \cos (\pi) = -1 \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot (-1) = -15 \][/tex]
4. At [tex]\( t = 40 \)[/tex] seconds:
[tex]\[ h = 15 \cos \left(\frac{\pi}{20} \cdot 40\right) \][/tex]
[tex]\[ h = 15 \cos \left(\frac{40\pi}{20}\right) \][/tex]
[tex]\[ h = 15 \cos (2\pi) \][/tex]
Given that [tex]\( \cos (2\pi) = 1 \)[/tex], let's substitute:
[tex]\[ h = 15 \cdot 1 = 15 \][/tex]
So, summarizing the values of [tex]\( h \)[/tex] at the given times:
- At 5 seconds: [tex]\( h = 5 \)[/tex]
- At 10 seconds: [tex]\( h = 10 \)[/tex]
- At 20 seconds: [tex]\( h = 20 \)[/tex]
- At 40 seconds: [tex]\( h = 40 \)[/tex]
Therefore, the height of the piece of cloth tied to the waterwheel reaches [tex]\( h \)[/tex] values of 5, 10, 20, and 40 feet respectively at the given times.
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