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Answer:
Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.
Explanation:
The power delivered by a battery is given by the formula:
P = VI
where,
P = Power Delivered by battery in 1 second
V = Voltage of battery = 12 volt
I = Current stored in battery
But, if we multiply both sides of equation by time (t), then:
Pt = VIt
where,
Pt = Power x Time = E = Energy Stored = ?
It = Rating of Battery = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
<u>E = 36000 J = 36 KJ</u>
<u>Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.</u>
Quasi frequency = 4√6
Quasi period = π√6/12
t ≈ 0.4045
<u>Explanation:</u>
Given:
Mass, m = 20g
τ = 400 dyn.s/cm
k = 3920
u(0) = 2
u'(0) = 0
General differential equation:
mu" + τu' + ku = 0
Replacing the variables with the known value:
20u" + 400u' + 3920u = 0
Divide each side by 20
u" + 20u' + 196u = 0
Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.
r² + 20r + 196 = 0
Determining the roots:
![r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B-20%20%2B-%20%5Csqrt%7B%2820%29%5E2%20-%204%281%29%28196%29%7D%20%7D%7B2%281%29%7D)
r = -10 ± 4√6i
The general solution for two complex roots are:
y = c₁ eᵃt cosbt + c₂ eᵃt sinbt
with a the real part of the roots and b be the imaginary part of the roots.
Since, a = -10 and b = 4√6
u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t
u(0) = 2
u'(0) = 0
(b)
Quasi frequency:
μ = ![\frac{\sqrt{4km - y^2} }{2m}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B4km%20-%20y%5E2%7D%20%7D%7B2m%7D)
![= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Csqrt%7B4%283929%29%2820%29%20-%20%28400%29%5E2%7D%20%7D%7B2%2820%29%7D%20%5C%5C%5C%5C%3D%204%5Csqrt%7B6%7D)
(c)
Quasi period:
T = 2π / μ
![T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6} }{12}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20%7D%7B4%5Csqrt%7B6%7D%20%7D%20%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B%5Cpi%5Csqrt%7B6%7D%20%20%7D%7B12%7D)
(d)
|u(t)| < 0.05 cm
u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05
solving for t:
τ = t ≈ 0.4045
Answer:
In a voltaic cell, current is produced by connecting an oxidation reaction half cell to a reduction reaction half cell in an electrolyte solution. Oxidation and reduction reactions (redox reactions) are chemical reactions involving a transfer of electrons from one atom in the reaction to another. When two different oxidation or reduction reactions are connected electrically by connecting the cathode to the anode, a current is formed. The direction depends on the type of reaction taking place at the terminal.
The first step would be to determine metals to be used as the cathode and the anode.