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
Answer:
The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.
Explanation:
The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform. In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
