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Gnom [1K]
2 years ago
15

What is international system of units?​

Physics
2 answers:
Gnom [1K]2 years ago
7 0

The International System of Units (abbreviated SI from systeme internationale, the French version of the name) is a scientific method of expressing the magnitudes or quantities of important natural phenomena.

Pavel [41]2 years ago
4 0
The International System of Units, known by the international abbreviation SI in all languages and sometimes pleonastically as the SI system, is the modern form of the metric system and the world's most widely used system of measurement.
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A compound is discovered and its molecular weight is determined to be 526g/mol. however, one of the elements has not yet been id
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Sulphur

The missing element is sulphur

Here the compound is composed of  element X and chlorine.

it is given,

XCl (6) ----> 6Cl

mass = 13.1%

it is X and Cl = 100%-13.10%

                     = 86.90%

chlorine = 86.90            

X = 13.10

We assume 100g of sample

so according to the above solved data, in 100g of sample we have 86.90 g of chlorine.

now we split that chlorine into two moles of chlorine.

  • Every mole of sale has 35.45 grams of sales. 2.451 moles of seal are mine after that.
  • We are aware that cl is six times more than X. To find the moles of X, I must divide this number by six.
  • we discover that we have  0.4086 moles of X when we divide this by six.
  • We obtain those moles of X from 13.10 g of X.  
  • Given that Mueller masses grams per mole, I can compute the molar mass.
  • 13.10 grams of X are contained in 0.4086 moles of X.
  • Or, to put it another way, each remote has a molar mass of 32–06 grams.
  • The element with this molar mass is sulfur, as I can see from the periodic table.
  • Okay, so the element we're looking for in this situation is sulfur.

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7 0
1 year ago
a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

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v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

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Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

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