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erica [24]
3 years ago
13

According to John Dalton’s observations when elements combine in a compound

Physics
2 answers:
Andrej [43]3 years ago
6 0

<u>Answer:</u> When elements combine in a compound, the ratio of their masses is always the same

<u>Explanation:</u>

Dalton's theory is based on mainly two laws:

  • Law of conservation of mass
  • Law of constant composition.

Law of conservation of mass states that mass can neither be created nor be destroyed, but it can transformed from one form to another form.

Law of constant composition states that a compound always contain the elements in the fixed ratio by their masses.

<u>For Example:</u> In water (H_2O), the hydrogen and oxygen are present in the fixed ratio of 1 : 9 by their mass.

Hence, when elements combine in a compound, the ratio of their masses is always the same

gayaneshka [121]3 years ago
3 0

The answer is A.) The ratio of thier masses is always the same

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.
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The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

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1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

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kolbaska11 [484]

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A man walking on a tightrope carries a long a pole which has heavy items attached to the two ends. If he were to walk the tight-
katen-ka-za [31]

Answer:

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this value is much larger and with it it is easier to restore balance.I

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Therefore the moment of the masses and the open is the one that must be zero.

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As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.

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