Answer:
60 Ω
Explanation:
R(com) = 15 Ω
1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn
1/15 = 1/20 + 1/R2
1/R2 = 1/15 - 1/20
1/R2 = (4 - 3) / 60
1/R2 = 1/60
R2 = 60 Ω
así, la combinada de resistencia necesaria es 60 Ω
The mass of a substance is given in atomic mass units and is calculated by adding the average atomic masses of all the atoms in the substance's chemical formula.
<h3>What empirical formula represents the total average atomic mass of every atom?</h3>
The Method The average atomic masses of all the atoms included in a formula's representation are added to get the mass of any molecule, formula unit, or ion. It has no bearing on the number of significant figures because the number of atoms is an exact quantity. One H2O molecule weighs 18.02 amu on average.
<h3>What connection exists between the empirical formula and the molecular formula?</h3>
You can determine the number of atoms of each element in a molecule using its molecular formula. These empirical formulations provide the most basic or reduced elemental ratio of a compound. The empirical formula and the molecular formula of a substance are same if the molecular formula can no longer be decreased.
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<span>Niche - </span><span>An organism's particular role in an ecosystem, or how it makes its living.</span>
Answer:
X = 6910319.7 m
Explanation:
let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.
on the surface of the earth, the gravitational acceleration is given by:
g1 = GM/(r^2) = [(6.67408×10^-11)(5.972×10^24)]/[(6371×10^3)^2] = 9.82 m/s^2
at X meters above the earth's surface, g2 = 85/100(9.82) = 8.35m/s^2
then:
g2 = GM/(X^2)
X^2 = GM/g2
X = \sqrt{GM/g2}
= \sqrt{(6.67408×10^-11)(5.972×10^24)/ 8.35
= 6910319.7 m
Therefore, the acceleration of gravity becomes 85% of what it is on the surface at 6910319.7 m .
The value of the spring constant of the spring is 14.7 N/m.
The given parameters;
- <em>mass attached to the spring, m = 30 g</em>
- <em>length of the spring, L₀ = 10 cm</em>
- <em>angular speed of the mass = 90 rpm</em>
- <em>final length of the spring, L₁ = 12 cm</em>
The extension of the spring is calculated as follows;
The value of the spring constant of the spring is calculated by applying Hooke's law;
Learn more about Hooke's law here:brainly.com/question/4404276
