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worty [1.4K]
3 years ago
12

Why must the height of the meniscus in the graduated cylinder match the height of the water in the tub when measuring volume?

Physics
1 answer:
galben [10]3 years ago
8 0

Answer:

See explanation

Explanation:

First, in order for you to understand, remember the basic concept of meniscus in graduated cylinder.

<em>"The meniscus is the curve seen at the top of a liquid in response to its container. The meniscus can be either concave or convex, depending on the surface tension of the liquid and its adhesion to the wall of the container".</em>

Now, according to this definition, and for water, the reading of the volume must be donde at the bottom of the curve of the meniscus. This is because the water  gives a concave curve.

If you read it and matches the height of water, you are getting two results:

One, get an accurate value or volume, because it's been done at eye level.

The second fact is that when you do the reading this way, The total pressure is made equal to the atmospheric pressure by adjusting the height of the cylinder until the water level is equal.

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<span>the same amount of work being done over a longer period of time.</span>
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S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to
denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

5 0
1 year ago
which of these changes does a submarine encounter as it returns from the bottom of the ocean to the surface of the ocean
fiasKO [112]

a.the amount of sunlight increases.

Explanation:

As a submarine rises to the surface, the change it encounters that is true from the given options is that the amount of sunlight increases.

The bottom of the ocean is dark and receives little to no sunlight due to the scattering of the rays by ocean water.

  • As the submarine rises, the volume of water column on it decreases and the pressure on it decreases too.
  • Also, the temperature rises steadily to the surface.

learn more:

Heat and temperature brainly.com/question/914750

#learnwithBrainly

7 0
3 years ago
What is the period of a soundwave whose wavelength is 20.0 m? Use values from the book and show ALL of your work.
Liono4ka [1.6K]
<h3><u>Answer;</u></h3>

Period = 1/17 seconds

<h3><u>Explanation;</u></h3>
  • Wavelength is related to period by the expression:

<em>speed = wavelength / period </em>

  • If we are given the speed, then we can easily calculate the period at the wavelength of 20 m.

<em>Given the speed of sound wave as 340 m/s </em>

<em>Period = Wavelength/ speed</em>

<em>            = 20 m/340 m/s</em>

<em>            </em><u><em>= 1/17 seconds</em></u>

7 0
2 years ago
Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?
KATRIN_1 [288]

Answer:

  0.83 g/cm³

Explanation:

The volume of the alcohol is ...

  (33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

  (9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

  42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

  33.2 g + 9 g = 42.4 g

So, the resulting density is ...

  (42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

  (33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

3 0
2 years ago
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