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White raven [17]
2 years ago
9

Q C Blaise Pascal duplicated Torricelli's barometer using a red Bordeaux wine, of density 984kg/ m³ , as the working liquid (Fig

. P14.16). (a) What was the height h of the wine column for normal atmospheric pressure?
Physics
1 answer:
levacccp [35]2 years ago
6 0

The pressure within Earth's atmosphere is referred to as atmospheric pressure or barometric pressure. 10.701 was the height h of the wine column for normal atmospheric pressure.

<h3>What is Atmospheric pressure?</h3>

The pressure within Earth's atmosphere is referred to as atmospheric pressure or barometric pressure. The definition of the standard atmosphere is 101,325 Pa, or the same as 1013.25 millibars, 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.

The force per unit area that an atmospheric column exerts is known as atmospheric pressure, often known as barometric pressure (that is, the entire body of air above the specified area). A weather indicator is atmospheric pressure. There will typically be clouds, wind, and precipitation when a low-pressure system enters a region. Fair, quiet weather is frequently a result of high pressure systems.

We have the density for the red Bordeaux wine given $\rho=965 \frac{\mathrm{kg}}{\mathrm{m}^3}$, the atmospheric pressure on the Torricelli's barometer is given by:

$$P_{a t m}=\rho g h$$

Solving for the height of wine in the column we have this:

$h=\frac{P_{\text {anm }}}{\rho g}$

And replacing we have:

h=\frac{101300 \mathrm{~Pa}}{965 \frac{\mathrm{kg}}{\mathrm{m}^3 9.81 \frac{\mathrm{m}}{\mathrm{m}^2}}}=10.701 \mathrm{~m}$$

So the height of the red Bordeaux wine would be $\mathrm{h}=10.701 \mathrm{~m}$. A very high value on this case if we compare with the usual values for this variable.

To learn more about Atmospheric pressure refer to:

brainly.com/question/19587559

#SPJ4

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Answer:

1.41 m/s, 7.85 rad/s

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We can start by calculating the tangential velocity, which is given by:

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T = 0.8 s is the period

Substituting,

v=\frac{2\pi(0.18)}{0.8}=1.41 m/s

Now we can also calculate the angular velocity,  which is given by:

\omega=\frac{2\pi}{T}

where again,

T = 0.8 s is the period

Substituting,

\omega=\frac{2\pi}{0.8}=7.85 rad/s

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Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and fricti
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Answer:

q_1 = \pm 1.68 \times 10^{-7} C

q_2 = \pm 6.68 \times 10^{-7} C

Explanation:

As we know that the force between two small spheres is given as

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here we know that

q_1 , q_2 = charges on two small spheres

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now the force between them is given as

0.045 = \frac{(9\times 10^9)(q_1q_2)}{0.15^2}

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now when two spheres are connected together then the charge on them is equally divided

q = \frac{q_1+q_2}{2}

now the force between them is given as

F = \frac{k(\frac{q_1+q_2}{2})^2}{0.15^2}

0.070 = \frac{(9\times 10^9)(\frac{q_1+q_2}{2})^2}{0.15^2}

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q_1 = \pm 1.68 \times 10^{-7} C

q_2 = \pm 6.68 \times 10^{-7} C

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