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ipn [44]
3 years ago
10

An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t

he local atmospheric pressure) in kPa inside of the diving chamber required to keep water from entering the chamber. (Assume SG=1.03)
Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

P(z)=P_{0}+\rho _wgh

Now the gauge pressure is given by

P(z)-P_{0}=\rho _wgh

Applying values we get

P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa

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Answer:

Explanation:

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10log\frac{I}{I_0} = Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

10log\frac{I}{10^{-12}} = 71

log\frac{I}{10^{-12}} = 7.1

\frac{I}{10^{-12}} = 10^{7.1}

I= 10^{-4.9} W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x 10^{5.4}

I= 10^{-6.6 } W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

I= 10^{-4.9}  = 2π² x 504²xA²x 1.21 x  346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

10^{-6.6} = 2π² x 504²xA²x 1.21 x  346

A² = 1.1978 x 10⁻¹⁶

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A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87° from the direction of th
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Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, v=8.2\times 10^3\ m/s

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Magnetic force, F = 0.002 N

The magnetic force is given by :

F=qvB\ sin\theta

B is the magnetic field  

B=\dfrac{F}{qv\ sin\theta}

B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}

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