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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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Answer:

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Answer:

2(maximum), -2(minimum), -2(maximum).

Explanation:

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Applying the condition given, we have s(0)= 0.

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Which means that; 0+C= 0. That is; C=0.

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Take note that;  sin(π/2) = sin(5π/2) = 1 .

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Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typica
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Answer:

The break even cost is $0.0063825

Explanation:

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Wattage = W

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Discount rate = 11%

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Price of bulb = $0.39

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Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

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PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

<u>Calculate the EAC of bulb</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, <em>consider this equation 1</em>

<u>Calculate the EAC of CFL</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, <em>consider this equation 2</em>

<u>Equate 1 and 2 to find the amount of C</u>

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

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Answer:

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