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valentinak56 [21]
3 years ago
11

Legend has it that, many centuries ago,Archimedes jumped out of his bathtub and ran across town nakedscreaming "Eureka!" after h

e solved an especially difficultproblem. Though you may not have thought of things this way before,when you drink a glass of water, the water that you are drinkingcontains some water molecules that were in Archimedes' bathwaterthat day, because water doesn't get created or destroyed on a largescale. It follows the water cycle, which includes rain,evaporation, flowing of rivers into the ocean, and so on. In themore than two thousand years since his discovery, the watermolecules from Archimedes' bathwater have been through this cycleenough times that they are probably about evenly distributedthroughout all the water on the earth. When you buy a can of soda,about how many molecules from that famous bathtub of Archimedes arethere in that can?
Physics
1 answer:
docker41 [41]3 years ago
3 0

Answer: 8652.36857143 x 10^{4} molecules

Explanation:

We need a lot of assumptions in this case.

1. We don't know the mass of water that was present in Archimedes' bath tub that day.

2. Since it recycles, we don't know the total mass of water present on earth.

Assuming the total mass of water on earth is 1.4 × 10^{18} tonnes;

converting tonnes to kgs -->>  1.4 × 10^{21}kg

Since only 2.5% of this is fresh, (the rest is saline and ocean-based)

-->> 2.5% of 1.4 × 10^{21}kg = 0.035  × 10^{21}kg

An average bath tub holds about 80 gallons of water.

80 gallons to kgs = 302.8329 kg

Using Avogadro's number,

we have about 6 × 10^{23} molecules in 18g of water;

1kg of water would contain about 3.3 × 10^{25} molecules.

An average water can holds about 355mL of fluid.

Therefore the number of molecules in the can is 10^{25}.

The number of bathtub molecules present in the can is;

302.8329 x 10^{25} / 0.035 x 10^{21}

= 8652.36857143 x 10^{4} molecules

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Answer:

We have learned that refraction occurs as light passes across the boundary between two media. Refraction is merely one of several possible boundary behaviors by which a light wave could behave when it encounters a new medium or an obstacle in its path.

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3 years ago
Please send me solution of the question pls​
ELEN [110]

Answer:

20m

Explanation:

P.E=mgh

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Divide both side by 100

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3 years ago
The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the
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To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

V_d = \frac{I}{nAq}

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

I = 25 A

A= 1.2*20 *10^{-6} m^2

q= 1.6*10^{-19}C

N = 8.47*10^{19} mm^{-3}

V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}

V_d = 7.68*10^{-5}m/s

The hall voltage is given by

V=\frac{IB}{ned}

Where

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d = distance

e = charge of electron

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3 years ago
What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are
nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is 15.58 g/cm^3.

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = 19.32g/cm^3

The density of silver = 10.1g/cm^3

The density of copper =8.8g/cm^3

Volume=\frac{Mass}{Density}

Volume of the gold in the mixture = V_1=\frac{18x}{19.32 g/cm^3}

Volume of the silver in the mixture = V_2=\frac{5x}{10.1 g/cm^3}

Volume of the copper in the mixture = V_3=\frac{1x}{8.8 g/cm^3}

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

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8 0
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Answer:

C

Explanation:

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