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Vesnalui [34]
3 years ago
9

A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a poin

t south of the lake. From that point to the left side of the lake, the distance is 105 meters. From that point to the right side of the lake is 119 meters. The angle between these two measurements is 83 degrees. How far is it across the lake?
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

149 m

Explanation:

The distances across the lake is forming a triangle.  

let the distance between the point and the left side be 'x'

and the distance between the point and the right be 'y'

and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:

∠Z = 83°

x = 105 m

y = 119 m

Now, applying the Law of Cosines, we get  

z² = x² + y² - 2xycos(Z)  

Substituting the values in the above equation, we get

z² = 105² + 119² - 2×105×119×cos(83°)

or

z = √22140.48

or

z = 148.796 m ≈ 149 m

The point is 149 m across the lake

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Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

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By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

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Here

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  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

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