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Blizzard [7]
3 years ago
13

If an air traffic controller is describing the speed of an airplane as well as the direction in which it should begin its decent

, then the air traffic controller is describing the airplane's
a. magnitude.
b. velocity.
c. speed.
d. acceleration.
Physics
2 answers:
Sever21 [200]3 years ago
8 0
I think the correct answer from the choices listed above is option B. The air traffic controller is describing the airplane's velocity. Velocity<span> is a physical vector quantity; both magnitude and direction are needed to define it. Hope this answers the question. Have a nice day.</span>
liberstina [14]3 years ago
5 0

The correct answer for this is B. Velocity. To find velocity you need the speed and direction, what we have. Sorry for the wait, Good luck :)

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Choose the 200 kg refrigerator. Set the applied force to 400 N (to the right). Be sure friction is turned off.What is the net fo
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So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

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3 0
3 years ago
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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
Zanzabum

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

S_1 O=4m

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

The intensity at O when both speakers are on is given by

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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