Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
Answer:
We know the information about atomic size, energy, electronic configuration etc. of atom from the periodic table.
Explanation:
- Periodic table is the arrangement of elements that are arranged according to their properties and electronic configuration.
- In periodic table, on furthest right side of the periodic table, noble gases like He, Ne, Ar etc are arranged.
- The atomic number of element increases while moving from left towards right in the periodic table.
- The metallic character of element decreases as we proceed the table towards right.
- They readily accept electron to fill the valence shell hence becoming more metallic in character.
The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.
Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
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Answer:
Work done = = 5 kJ
Explanation:
Given data:
volume of nitrogen 
Polytropic exponent n = 1.4
![\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%20%3D%20%5B%5Cfrac%7BP_2%7D%7BP_1%7D%5D%5E%7B%5Cfrac%7Bn-1%7D%7Bn%7D)
putting all value
![\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7B473%7D%20%3D%20%5B%5Cfrac%7B80%7D%7B150%7D%5D%5E%7B%5Cfrac%7B1.4-1%7D%7B1.4%7D)
polytropic process is given as
work done 
= 5 kJ
