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3 years ago

5

What is an anemometer used to measure?

1 answer:

wariber [46]3 years ago

8 0

An anemometer is used to measure wind speed. It is commonly found in weather stations.

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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

5 0

3 years ago

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

3 years ago

A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that

AleksAgata [21]

Answer:

a = 2.275 10⁻⁴ m

Explanation:

This is a diffraction problem that is described by the equation

a sin θ = m λ

The first dark minimum occurs for m = 1

a = λ / sin θ

The angle can be found by trigonometry,

tan θ = y / x

θ = tan⁻¹ y / x

Let's reduce the magnitudes to the SI system

y = 8.24 mm = 8.24 10⁻³ m

λ = 625 nm = 625 10⁻⁹ m

θ = tan⁻¹ 8.24 10⁻³ / 3.00

θ = 0.002747 rad

We calculate

a = 625 10⁻⁹ / sin 0.002747

a = 2.275 10⁻⁴ m

8 0

3 years ago

A basketball player can leap upward 0.42 m. What is the magnitude of his initial velocity at the start of the leap? Use g=9.81 m

Tatiana [17]

Farmula 2gh=vf²-vi²
2×0.42×9.81=0²-vi²
8.2404=vi²
Vi=✓8.2404
Vi=2.87m/sec

8 0

3 years ago

63°C = _____ 63 K 336 K 210 K 163 K

Nata [24]

Is is the second one. 336 k. 273.15K= 0C.

3 0

3 years ago

Read 2 more answers