Velocity is distance over time

. If it wants it in km/h you're good to go otherwise if its m/s you need to convert km to m by multiplying 30 by 1000 and t in hours to seconds by multiplying 0.5 by 3600.
The appropriate response is amplitude. A measure of its change over a solitary period. There are different meanings of plentifulness, which are all elements of the extent of the distinction between the variable's outrageous qualities. In more seasoned writings, the stage is now and again called the adequacy.
Answer:
Approximately
.
Explanation:
Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (
) is equal to
.
There are two half-reactions in this question.
and
. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of
should be positive.
In this case,
is positive only if
is the reaction takes place at the cathode. The net reaction would be
.
Its cell potential would be equal to
.
The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:
,
where
is the number moles of electrons transferred for each mole of the reaction. In this case the value of
is
as in the half-reactions.
is Faraday's Constant (approximately
.)
.
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