PLEASE HELP WILL GIVE BRAINLIEST In an experiment, the hypothesis is that if leaf color is related to temperature, then exposing
2 answers:
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The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
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
as in the half-reactions.
is Faraday's Constant (approximately
.)
.