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3 years ago

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A cyclist maintains a constant velocity of 4.1 m/s headed away from point A. At some initial time, the cyclist is 244 m from poi

nt A. What will be his displacement from his starting position after 60 s?

1 answer:

astra-53 [7]3 years ago

6 0

Answer:

$d=490\ m$ is his final displacement from the point A after 60 seconds.

Explanation:

Given:

Cyclist is moving away from A.

velocity of cyclist, $v=4.1\ m.s^{-1}$

displacement of the cyclist from point A at the time of observation, $d_i=244\ m$

time after which the next observation is to be recorded, $t=60\ s$

Now as the cyclist is moving away from point A his change in displacement after the mentioned time:

$\Delta d=v.t$

$\Delta d = 4.1 \times 60$

$\Delta d=246\ m$

<u>Now the the final displacement from point A after the mentioned time:</u>

$d=d_i+\Delta d$

$d=244+246$

$d=490\ m$

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Answer:

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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

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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

$S_1 S_2$ =3m

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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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$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

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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