Answer:
Explanation:
![\mathbf{F(A, B, C) = \bar A \bar B C + B \bar C + AB}](https://tex.z-dn.net/?f=%5Cmathbf%7BF%28A%2C%20B%2C%20C%29%20%20%3D%20%5Cbar%20A%20%5Cbar%20B%20C%20%2B%20B%20%5Cbar%20%20C%20%2B%20AB%7D)
![\mathbf{F(A, B, C) = \bar A \bar B C + AB \bar C + \bar A B \bar C + ABC + AB \bar C}](https://tex.z-dn.net/?f=%5Cmathbf%7BF%28A%2C%20B%2C%20C%29%20%20%3D%20%5Cbar%20A%20%5Cbar%20B%20C%20%2B%20AB%20%20%5Cbar%20%20C%20%2B%20%5Cbar%20A%20B%20%20%5Cbar%20C%20%2B%20ABC%20%2B%20AB%20%20%5Cbar%20C%7D)
![\mathbf{F(A, B, C) = 001,110,010,111,110}](https://tex.z-dn.net/?f=%5Cmathbf%7BF%28A%2C%20B%2C%20C%29%20%20%3D%20001%2C110%2C010%2C111%2C110%7D)
Hence;
![\mathbf{F(A, B, C) = \sum m (1,2,6.7)}](https://tex.z-dn.net/?f=%5Cmathbf%7BF%28A%2C%20B%2C%20C%29%20%20%3D%20%5Csum%20m%20%281%2C2%2C6.7%29%7D)
![\mathbf{ A \ \ \ B \ \ \ C \ \ \ \ \ \ \ F } \\ \\ \mathbf{ 0 \ \ \ \ 0 \ \ \ \ 0 \ \ \ \ \ \ \ 0 } \\ \\ \mathbf{ 0 \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ \ \ \ 1 }\to \ \ D_0 = C \\ \\ \mathbf{ 0 \ \ \ \ 1 \ \ \ \ 0 \ \ \ \ \ \ \ 1 } \to \ \ D_1 = \bar C \\ \\ \mathbf{ 0 \ \ \ \ 1 \ \ \ \ 1 \ \ \ \ \ \ \ 0 } \\ \\ \mathbf{ 1 \ \ \ \ 0 \ \ \ \ 0 \ \ \ \ \ \ \ 0 } \\ \\ \mathbf{ 1 \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ \ \ \ 0 }](https://tex.z-dn.net/?f=%5Cmathbf%7B%20A%20%20%5C%20%5C%20%5C%20%20B%20%5C%20%5C%20%5C%20C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20F%20%7D%20%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%200%20%20%5C%20%5C%20%20%5C%20%5C%20%200%20%5C%20%5C%20%20%5C%20%5C%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%200%20%7D%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%200%20%20%5C%20%5C%20%20%5C%20%5C%20%200%20%5C%20%5C%20%20%5C%20%5C%201%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%201%20%7D%5Cto%20%20%20%5C%20%5C%20D_0%20%3D%20C%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%200%20%20%5C%20%5C%20%20%5C%20%5C%20%201%20%5C%20%5C%20%20%5C%20%5C%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%201%20%7D%20%5Cto%20%5C%20%5C%20D_1%20%3D%20%5Cbar%20C%20%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%200%20%20%5C%20%5C%20%20%5C%20%5C%20%201%20%5C%20%5C%20%20%5C%20%5C%201%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%200%20%7D%20%20%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%201%20%20%5C%20%5C%20%20%5C%20%5C%20%200%20%5C%20%5C%20%20%5C%20%5C%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%200%20%7D%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%201%20%20%5C%20%5C%20%20%5C%20%5C%20%200%20%5C%20%5C%20%20%5C%20%5C%201%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%200%20%7D)
![\\ \\ \mathbf{ 1 \ \ \ \ 1 \ \ \ \ 0 \ \ \ \ \ \ \ 0 } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \to D_3 = 1 \ \ ; D_2 = 0 \\ \\ \mathbf{ 1 \ \ \ \ 1 \ \ \ \ 1 \ \ \ \ \ \ \ 1 }](https://tex.z-dn.net/?f=%5C%5C%20%5C%5C%20%5Cmathbf%7B%201%20%20%5C%20%5C%20%20%5C%20%5C%20%201%20%5C%20%5C%20%20%5C%20%5C%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%200%20%7D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5Cto%20%20D_3%20%3D%201%20%5C%20%5C%20%3B%20D_2%20%3D%200%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%201%20%20%5C%20%5C%20%20%5C%20%5C%20%201%20%5C%20%5C%20%20%5C%20%5C%201%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%201%20%7D)
The 4-to-1 MUX and inverter is shown in the attached file below.
Answer:
Magnetic fields can apply forces forces by the interaction between magnets and magnetic material
Explanation:
In a maglev train is made up of a train cart with superconducting magnets beneath, which is then mounted above specially designed maglev guideway. The non-contact motion of the train by the magnets is due to interaction between the magnetic field of the train with metallic loops in the guideway thereby inducing an opposing magnetic field in the loop which serve the purposes of 1. Keeping the train 5 inches above the guideway, 2. Maintaining horizontal stability and 3. Movement of the train through alternating the attractive and repulsive forces such that the resultant effect is the net movement of the train.
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-![\int\limits^a_b {P} \, dV](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7BP%7D%20%5C%2C%20dV)
a=![v_2\\](https://tex.z-dn.net/?f=v_2%5C%5C)
b=![v_1](https://tex.z-dn.net/?f=v_1)
from virial equation
![\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\](https://tex.z-dn.net/?f=%5Cfrac%7BPV%7D%7BRT%7D%3Dz%3D1%2B%5Cfrac%7BB%7D%7BV%7D%2B%5Cfrac%7BC%7D%7BV%5E2%7D%5C%5C%20%20%20%5C%5CP%3DRT%281%2B%5Cfrac%7BB%7D%7BV%7D%20%2B%5Cfrac%7BC%7D%7BV%5E2%7D%29%5Cfrac%7B1%7D%7BV%7D%5C%5C)
And
![W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV](https://tex.z-dn.net/?f=W%3D-%5Cint%5Climits%5Ea_b%20%7BRT%281%2B%5Cfrac%7BB%7D%7BV%7D%20%2B%5Cfrac%7BC%7D%7BV%5E2%7D%5Cfrac%7B1%7D%7BV%7D%20%20%7D%20%5C%2C%20dV)
a=![v_2\\](https://tex.z-dn.net/?f=v_2%5C%5C)
b=![v_1](https://tex.z-dn.net/?f=v_1)
Now calculate V1 and V2 at given condition
![\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}](https://tex.z-dn.net/?f=%5Cfrac%7BP1V1%7D%7BRT%7D%20%3D%201%2B%5Cfrac%7BB%7D%7Bv_1%7D%20%2B%5Cfrac%7BC%7D%7Bv_1%5E2%7D)
Substitute given values
= 1 x 10^5 , T = 373.15 and given values of coefficients we get
![10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2](https://tex.z-dn.net/?f=10%5E5%28v_1%29%2F8.314%2A373.15%3D1-242.5%2Fv_1%2B25200%2Fv_1%5E2)
Solve for V1 by iterative or alternative cubic equation solver we get
![v_1=30780 cm^3/mol](https://tex.z-dn.net/?f=v_1%3D30780%20cm%5E3%2Fmol)
Similarly solve for state 2 at P2 = 50 bar we get
![v_1=241.33 cm^3/mol](https://tex.z-dn.net/?f=v_1%3D241.33%20cm%5E3%2Fmol)
Now
![W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV](https://tex.z-dn.net/?f=W%3D-%5Cint%5Climits%5Ea_b%20%7BRT%281%2B%5Cfrac%7BB%7D%7BV%7D%20%2B%5Cfrac%7BC%7D%7BV%5E2%7D%5Cfrac%7B1%7D%7BV%7D%20%20%7D%20%5C%2C%20dV)
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
![W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP](https://tex.z-dn.net/?f=W%3D-%5Cint%5Climits%5Ea_b%20%7BP%28RT%28-1%2Fp%5E2%2B0%2BC%27%29%7D%20%5C%2C%20dP)
a=![v_2\\](https://tex.z-dn.net/?f=v_2%5C%5C)
b=![v_1](https://tex.z-dn.net/?f=v_1)
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
Answer:c
Explanation:
Because it will take out the fuel right away
Answer:
Explanation:
Given
charge is placed at ![x=0\ cm](https://tex.z-dn.net/?f=x%3D0%5C%20cm)
another charge of
is at ![x=3\ cm](https://tex.z-dn.net/?f=x%3D3%5C%20cm)
We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.
so net electric field is zero somewhere beyond negatively charged particle
Electric Field due to
at some distance r from it
![E_2=\frac{kq_2}{r^2}](https://tex.z-dn.net/?f=E_2%3D%5Cfrac%7Bkq_2%7D%7Br%5E2%7D)
Now Electric Field due to
is
![E_1=\frac{kq_1}{(3+r)^2}](https://tex.z-dn.net/?f=E_1%3D%5Cfrac%7Bkq_1%7D%7B%283%2Br%29%5E2%7D)
Now ![E_1+E_2=0](https://tex.z-dn.net/?f=E_1%2BE_2%3D0)
![\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ctimes%2011.5%7D%7B%28r%2B3%29%5E2%7D%5Cfrac%7Bk%5Ctimes%20%28-1.2%29%7D%7Br%5E2%7D%3D0)
![\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Br%7D%7Br%7D%3D%28%5Cfrac%7B11.5%7D%7B1.2%7D%29%5E%7B0.5%7D)
![\frac{3+r}{r}=3.095](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Br%7D%7Br%7D%3D3.095)
thus ![r=1.43\ cm](https://tex.z-dn.net/?f=r%3D1.43%5C%20cm)
Thus Electric field is zero at some distance r=1.43 cm right of