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3 years ago

The faintest sound an ear can hear (20 micro-pascals) is roughly ______ times smaller than atmospheric pressure at sea level.

Physics

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer: $5.06\times10^{9}$

At sea level, there is one standard atmospheric pressure which is equal to 101.325 kilopascals.

The pressure of faintest sound that a human ear can hear is 20 micro-pascals.

taking the ratio of two:

$\frac {101.325 kilopascals. }{20 micropascals} = \frac {101.325 \times 10^{3}Pa}{20 \times 10^{-6}Pa}$

$=5.06\times10^{9}$

Hence, the atmospheric pressure at sea level is $5.06\times10^{9}$ times greater than the faintest sound that a human ear can hear.

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An object is thrown into the air at 60m/s, straight up. What is its velocity at the highest point?

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Explanation:

Whenever an object is at its highest point, the velocity and acceleration of the object is zero.

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3 years ago

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

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Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

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For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

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3 years ago

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Answer:

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I got this answer from calculator soups physics calculators. I really recommend their website for formulas.

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3 years ago

WILL GIVE BRAINLIEST AND 30 POINTS!

mariarad [96]

Answer:

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Explanation:

Gallium is one such element used as a do/pant in a p-type semiconductor.

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.

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The central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

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The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

$F = \frac{M_e v^2}{r} \\\\$

where;

$M_e$ is mass of electron = 9.11 x 10⁻³¹ kg

$F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N$

Thus, the central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

Learn more about centripetal force here:brainly.com/question/20905151

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3 years ago