Explanation:
<u>Given</u><u>.</u> <u>Required</u><u> </u>
wo = 400N. MA=?
wi = 50N
<u>Solution</u>
MA = <u> </u><u>wo</u> = <u>400N</u> =8
wi. 50N
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
The temperature of the water getting colder would cause the liquid in the thermometer to drop due to less heat being transferred from the water to the liquid, so the liquid molecules are closer than when they have high energy.
C is the correct answer beacause it shows where it is happening in this cas “here”.
