Answer:
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Explanation:
Electric field due to plates
Ef = V/d
Ef = 2033 V / (2.0 * 10^-2 m )
Ef = 101650 V/m
So, we can write
Ef * q = m*g
q = m*g / E
f
The mass can be equal using the density and the volume so:
m = ρ * v
The volume can be find as:
v = 2.298 x 10 ⁻ ¹⁶ m³
q = ρ * v * g / Ef
q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Answer:
Tt = 70 + 135e^-0.031t
13 minutes
Explanation:
Given that :
Initial temperature, Ti = 205°
Temperature after 2.5 minutes = 195°
Temperature of room, Ts= 70
Using the relation :
Tt = Ts + Ce^-kt
Temperature after time, t
When freshly poured, t = 0
205 = 70 + Ce^-0k
205 = 70 + C
C = 205 - 70 = 135°
T after 2.5 minutes to find proportionality constant, k
Tt = Ts + Ce^-kt
195 = 70 + 135e^-2.5k
125 = 135e^-2.5k
125 / 135 = e^-2.5k
0.9259 = e^-2.5k
Take In of both sides :
−0.076989 = - 2.5k
k = −0.076989 / - 2.5
k = 0.031
Equation becomes :
Tt = 70 + 135e^-0.031t
t when Tt = 160
160 = 70 + 135e^-0.031k
90 = 135e^-0.031t
90/135 = e^-0.031t
0.6667 = e^-0.031t
In(0.6667) = - 0.031t
−0.405465 = - 0.031t
t = 0.405465/ 0.031
t = 13.071
t = 13 minutes
