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Marianna [84]
4 years ago
8

If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above a

tmospheric pressure to compress her bones by 0.130 % of their original volume?
Physics
1 answer:
Sever21 [200]4 years ago
8 0

This is an incomplete question, here is a complete question.

The bulk modulus for bone is 15.0 (GPa). If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by 0.130% of their original volume?

Answer : The change in pressure will be, 1.95\times 10^7Pa

Explanation : Given,

Bulk modulus = 15.0GPa=15.0\times 10^9Pa

Change in volume = 0.130 % of original volume

Let the original volume be, V

So, Change in volume = \frac{0.130}{100}\times V

Formula used for change in pressure is:

\Delta P=\beta \frac{\Delta V}{V}

Now put all the given values in this formula, we get:

\Delta P=(15.0\times 10^9Pa)\times \frac{(\frac{0.130}{100}\times V)}{V}

\Delta P=1.95\times 10^7Pa

Thus, the change in pressure will be, 1.95\times 10^7Pa

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The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2

For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2

The second distance, r₂, can be determined from sound intensity formula given as;

I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 =  \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 =   \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m

Therefore, the second distance of the sound from the source is 431.78 m.

7 0
3 years ago
1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of
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Answer:

The car will make the turn perfectly

Explanation:

Given that the centripetal force= mv^2/r

M= mass of the car

v = speed of the car

r= radius

Hence;

F = 1000 × (14)^2/50

F= 3920 N

The frictional force = μmg

μ = coefficient of static friction

m= mass

g = acceleration due to gravity

Frictional force= 0.6 × 1000× 10

Frictional force = 6000 N

The car will not skid off the curve because the frictional force is greater than the centripetal force.

6 0
3 years ago
a length 650 cm of thin thread wraps around a cylinder exactly 30 times calculate the circumference and the radius of the cylind
Eduardwww [97]

Answer:

circumference= 65/3  cm = 21.67 cm

radius R = 3.45 cm

Explanation:

To calculate the length of the circumference of the cylinder, we divide 650 cm by 30 (the number of times it wrapped exactly around it)

length of circumference= 65/3  cm = 21.67 cm

now use the formula of the circumference length to find the radius (R):

circumference length = 2 * pi * R

65/3 = 2 * pi * R

R = 65 / (6 pi)

R = 3.45 cm

5 0
3 years ago
You indicate that a symbol
Goryan [66]

Answer:

B. over the symbol.

Explanation:

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8 0
3 years ago
A machine does 1200 J of work in 1 min. What is the power developed
densk [106]

Answer:

20 watts

Explanation:

Big brain mode activated:

Power=1200J/60sec

Power=20 watts

8 0
3 years ago
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