Answer:
The answer is below
Explanation:
The main difference between a liquid and a gas is that when a liquid is under pressure, its volume "won't change apparently. The reason is that the distance between the molecules of a liquid is relatively small, and the molecules of a liquid extensively withstand the compressive forces. This is similar to the distance between the molecules of a solid."
Answer:
L/2.
Explanation:
Hola.
En este caso, dado que el hombre empuja el cilindro desde el punto L=0 y hasta la mitad de la longitud entre el hombre y el cilindro, es claro que el cilindro se desplaza un L/2 más de la distancia inicial entre el hombre y el cilindro, de este modo, inferimos que el hombre también ha caminado L/2.
¡Saludos!
<h3><u>Answer;</u></h3>
<u>An increase in pressure favors the formation of ozone </u>
<h3><u>Explanation;</u></h3>
- Ozone, O3, decomposes to molecular oxygen in the stratosphere according to the reaction
2O3(g) ⇆ 3O2 (g).
- There are more moles of product gas than moles of reactant gas. An increase in total pressure increases the partial pressure of each gas, shifting the equilibrium towards the reactants.
- Therefore; an increase in pressure favors backward reactions towards the formation of ozone.
It will rise, and then fall back to the Earth’s surface.
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
