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3 years ago

Determina el tamaño de la arista de un cubo hecho de plata, cuya masa es de 10.49 kg.

Physics

1 answer:

Sidana [21]3 years ago

7 0

Answer:

88.2 N

Explanation:

Datos

Lcubo = 10 cm = 0.1 m

Vcubo = Vfluido desalojado= 0.1 m x 0.1 m x 0.1 m = 10-3 m

mcubo = 10 kg

dfluido = 1000 kg/m3

g = 9.8 m/s2

Sabemos que el peso aparente de un cuerpo que se sumerge en un fluido es:

Paparente=Preal−Pfluido

Teniendo en cuenta que:

Preal = mcubo⋅gPfluido=E= dfluido⋅Vfluido⋅g

Como el cuerpo se sumerge completamente en el fluido, el volumen de fluido desalojado es exactamente el volumen del cubo. Por lo tanto si sustituimos los datos que nos proporcionan en el enunciado en la primera ecuación:

Paparente=mcubo⋅g−dfluido⋅Vfluido⋅g ⇒Paparente=10 kg ⋅9.8 m/s2 − 1000 kg/m3 ⋅10−3 m ⋅9.8 m/s2 ⇒Paparente = 88.2 N

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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

$S_1 O$ =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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