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Viktor [21]
3 years ago
14

Blaise Pascal duplicated Torricelli's barometer using a red Bordeaux wine of density 965 kg/m3 as the working liquid (see figure

below).
1. What was the height h of the wine column for normal atmospheric pressure?
2. Would you expect the vacuum above the column to be as good as that for mercury?
A. Yes
B. No
Physics
1 answer:
pogonyaev3 years ago
5 0

Answer:

a) h=10.701m

b) No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.

Explanation:

The Torricelli's experiment "was invented by the Italian scientist Evangelista Torricelli and the most important purpose of this experiment was to prove that the source of vacuum comes from atmospheric pressure"

Pressure is defined as "the force that is applied on any object in the direction perpendicular to the surface of the object in the unit area is known as the pressure. There are various types of pressure".

Part a

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

P_{atm}=\rho g h

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

h=\frac{P_{atm}}{\rho g}

And replacing we have:

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

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

Part b

No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.

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pychu [463]

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

7 0
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A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist
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Answer:

Explanation:

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= \frac{71\times10^3}{4 \times \pi\times(220)^2}

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

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The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and
Zolol [24]

Answer:

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T_f = 330.0 K

\eta = 7.07 mole

Explanation:

Part 1

below equation is used to determine the type Gas by determining \gamma value

\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

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T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)

substituing value to get final temperature

T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}

T_f = 330.0 K

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Answer:

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Explanation:

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