In a hydrolysis reaction between a strong acid and a weak base, the salt formed will have a pH less than 7. By virtue of this phenomenon between <span>weak bases and strong acids</span>, the anion of the strong acid will fail to attract the hydrogen ion<span>, while the cation from the weak base will donate a </span>proton<span> to the water forming a hydronium ion</span><span>.
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Answer:
1.2cm
Explanation:
V=(2ev/m)^1/2
=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2
=6.2x10^5m/s
Radius of resulting path= MV/qB
= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6
=0.012m
=1.2cm
Answer: The amplitude is 0. (assuming that the amplitude ot both initial waves is the same)
Explanation:
When two monochromatic light waves of the same wavelength and same amplitude undergo destructive interference, means that the peak of one of the waves coincides with the trough of the other, so the waves "cancel" each other in that point in space.
Then if two light waves undergo destructive interference, the amplitude of the resultant wave in that particular point is 0.
Answer:
Explanation:
Maximum force of friction possible = μmg
= .65 x 3.8 x 9.8
= 24.2 N
u = 72 x 1000 / 60 x 60
= 20 m /s
v² = u² - 2as
a = 20 x 20 / (2 x 30)
= 6.67 m / s²
force acting on it
= 3.8 x 6.67
= 25.346 N
Friction force possible is less .
So friction will not be able to prevent its slippage
It will slip off .
Answer:
a) x = ⅔ d
, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
