Answer:
Option D
Explanation:
As the question mentions losing its carbonation. Option D is the only option that mentions carbon dioxide gas leaving the solution.
Given Information:
Radius = ra = 2.60 cm = 0.026 m
Density = J = 15.0 nC/m
change in potential difference = ΔV = 200 V
Required Information:
Distance = d = ?
Answer:
distance = 0.088 m
Explanation:
As we know
ΔV = Vb - Va = J/4πε₀*ln(rb/ra)
Where ra and rb is the point where potential difference is Va and Vb respectively
1/4πε₀ = 9x10⁹ N.m²/C²
We want to find the distance d = rb - ra
ΔV = J/4πε₀*ln(rb/ra)
200 = 9x10⁹*15x10⁻⁹*ln(rb/ra)
200/135 = ln(rb/ra)
1.48 = ln(rb/ra)
taking e on both sides yields
e^(1.48) = rb/ra
4.39 = rb/ra
rb = 4.39*0.026
rb = 0.114 m
Therefore, the required distance is
d = rb - ra
d = 0.114 - 0.026
d = 0.088 m
Therefore, the other probe must be placed 0.088 m from the surface so that the voltmeter reads 200 V
Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1
