Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as
The current flowing through the inductor is given by
Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
![i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\](https://tex.z-dn.net/?f=i%28t%29%20%3D%20%5Cfrac%7B1%7D%7B0.005%7D%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%2C%2B%200%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5C%3A%20%5B%20%7B5%5C%3A%20%28t%20%2B%20%5Cfrac%7Be%5E%7B-0.5t%7D%7D%7B0.5%7D%29%5D_0%5Et%20%5C%5Ci%28t%29%20%3D%20200%5Ctimes5%5C%3A%20%5C%3A%20%5B%20%7B%20%28t%20%2B%202e%5E%7B-0.5t%7D%20%2B%202%20%29%5D%20%5C%5C)
So the current at t = 2 seconds is
The energy stored in the inductor at t = 2 seconds is
Acid mine drainage is the formation and movement of highly acidic water rich in heavy metals. This acidic water forms through the chemical reaction of surface water (rainwater, snowmelt, pond water) and shallow subsurface water with rocks that contain sulfur-bearing minerals, resulting in sulfuric acid.
Answer:
(a) 
(b) 
Explanation:
(a)
Volume, V of unit cell
Number of unit cells, N
Where 
is weight of material and 
is density of material
(b)
Number of atoms in paper clip
This is a product of number of unit cells and number of atoms per cell
Since iron has 2 atoms per cell
Number of atoms of iron=
Answer:
Explanation:
Step by step solved solution is given in the attached document.
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
For ideal gas, 
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
