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Physics

2 answers:

Veseljchak [2.6K]3 years ago

6 0

Answer:

If south-east Texas is where H then it is C, but if not the answer is A.

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

b or c pretty sure its b b tho

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What measurements or observations tells you that a car is accelerating

lutik1710 [3]

You need to observe the car at two different times.

-- The first time:
You write down the car's speed, and the direction it's pointing.

-- The second time:
You write down the car's speed and the direction it's pointing, again.

You take the data back to your lab to analyze it.

-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.

-- You compare the first and second direction. If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.

(Remember, "acceleration" doesn't mean "speeding up".
It means any change in speed or direction of motion.)

8 0

4 years ago

Read 2 more answers

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.

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