Answer:
Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body.
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = 
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = 
V = V₀
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀
0.1V₀ = V₀
0.1V₀/V₀ = 
0.1 = 
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
Each
ion contains three extra protons. Hence, the extra charge on each
=
C
Total charge = 0.035 pC
Total charge (Q) =
C
Let the number of
ions be n.
According to question:



n = 72917
Hence, the total number of ions needed to be transferred is 72917
The answer would be B.
(V=IR, 12-10i, 12/10=1.2)
I believe the answer is the fourth one, hope this helps