To solve this problem it is necessary to apply the equations related to the Force of Friction and Energy.
By definition the friction force is defined as
Where,
Frictional Constant
N = Normal Force -> mg
At the same time we have the definition of the Energy, which can be
E = F*v
Where,
Force
v = Velocity.
Then replacing with our values we have that,
F = 0.5 *34
F = 17
The energy then would be,
E = f v
E = 17 * 0.30 = 5.1 W
Therefore the rate at which heat is generated is 5.1W
The formula is F = ( q1 * q2 ) / r ^ 2
<span>where: q is the individual charges of each ion </span>
<span>r is the distance between the nuclei </span>
<span>The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. </span>
<span>From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. </span>
<span>For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. </span>
<span>So Far: ( BaO and SrO ) > ( NaF and KCl ) </span>
<span>The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. </span>
<span>For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. </span>
<span>Therefore: </span>
<span>Answer: SrO > BaO > NaF > KCl </span>
