<h3><u>Answer;</u></h3>
a) 178.6125 Joules
b) Work done does not depend on constant speed or acceleration up the ladder
<h3><u>Explanation</u>;</h3>
Work done =Fd times the cosine of angle between;
where F is force d is distance m is the mass a is acceleration
.F=ma. In this case a=g=9.81 m/s^2 and cos 30=0.866
Therefore;
W= 75 × 9.81 × 2.75 × .866
= 178.6125 Joules
b. The work done by the painter does not depend on whether the painter climbs at constant speed or accelerates up the ladder.
W=3376.945 Joules
Remember opposites attract and same charges repel each other.
Object A= negatively charged.
Object A and B attract so B must be positively charged.
Object B and C repel so because B is positively charged C must also be positively charged.
Object C and D attract and because C is positively charged, D must be negatively charged.
<span>This apparent measurement difference is due to a change in position of the observer, and is called "Parallax"
In short, Your Answer would be Option C
Hope this helps!</span>
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
