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3 years ago

5

Consider two tubes filled with water at the same height, one with fresh water and the other tube with salt water. The pressure i

s greater at the bottom of the tube with:
A.) fresh water
B.) salt water
C.) both the same
D.) depends on whether the tubes have the same cross-sectional area

1 answer:

Olegator [25]3 years ago

3 0

Answer:

B

Explanation:

The correct option for the question is B that is salt water. In salt water, the density of water is higher so the pressure at the end of tube containing salt water will be greater. As according to the hydrostatic law the pressure at a given point will be directly proportional to the distance travelled as well.

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Answer:

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Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

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2 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

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Answer:

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Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

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Divide equation (1) by equation (2), we get

tan θ = Fe / mg

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$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

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