General Assignment Help: How to define a portfolio of Outsourced information technologies services...

Q) How to define a portfolio of Outsourced information technologies services??

Sol: 

Portfolio of Outsourced Information Technologies services can be defined as identifying and understanding the customer requirements, also providing a way to communicate the value that the IT organization delivers to the business. The portfolio of IT services also allows the IT organization to say to the business, “Here is what we deliver to you”. Ultimately, it is the list of what customers ‘buy’ from the IT organization by paying the Organization.

Looking for the solutions of few mathematics questions?

a) ABC' + B C' D' + B C + C' D = B + C' D

Sol:

LHS = (ABC’+BC)+ (BC’D’+C’D)

       =B(C+C’A) +C’ (D+D’B)

       =B(C+A) +C’ (D+B)

       =AB+BC+C’D+C’B

      =AB+C’D+B (C+C’)                //C+C’=1

      =AB+C’D+B

      =B (1+A) +C’D                        //1+A=1

      =B+C’D

      =RHS

b) WY + W' Y Z' + WXZ + W' X Y' = WY + W' X Z' + X' Y Z' + X Y' Z

Sol:

RHS=wy + w’xz’ + x’yz’ + xy’z

        =WY(1+XZ)+ w’xz’(Y+Y’)+ x’yz’ (W+W’)+ xy’z(W+W’)         //Y+Y’=W+W’=1+XZ=1

        = wy + wxyz + w’xyz’ + w’xy’z’ +wx’yz’ + w’x’yz’ + wxy’z + w’xy’z

        =wy + w’xyz’ + w’x’yz’ + wxyz +wxy’z + w’xy’z + w’xy’z’ + wx’yz’

        =wy + wx’yz’ + w’yz’(x + x’) + wxz(y + y’)+w’xy’(z + z’)

        =wy(1 + x’z’) + w’yz’ + wxz + w’xy’

        = WY + W' Y Z' + WXZ + W' X Y'

        =LHS

c) A D' + A' B + C'D + B'C = (A' + B' + C' + D')(A + B + C + D') = AC’+A’B +B‘C + D’

Sol:

RHS=(A' + B' + C' + D')(A + B + C + D')

        =A’A+A’B+A’C+A’D’+B’A+B’B+B’C+B’D’+C’A+C’B+C’C+C’D’+D’A+D’B+D’C+D’D’

        =A’B+A’C+A’D’+B’A+B’C+B’D’+C’A+C’B+C’D’+D’A+D’B+D’C+D’   //AA’=0; D’D’=D’

        = AC’+A’B +B‘C + D’+ A’C (B+B’) + B’A (C+C’) + C’B (A+A’)    //A+A’=1

        = AC’ (1 + B’ + B) + A’B (1 + C’ + C) +B’C(1 + A + A’) + D’    //1+A=1

        = AC’+A’B +B‘C + D’

        =LHS

Write-Ups Assignment Help: JIT logistics/supply chain management for what...

Q) JIT logistics/supply chain management for what (now defunct) America automaker.

Sol:

JUST-IN-TIME LOGISTICS SUPPORT IS USED FOR THE AUTOMOBILE INDUSTRY since JIT drastically reduces the investment as well as also the total cost of the operation. In the early 1970s the United States automobile industry realized that the advantage of adopting the JIT would bring the US automakers where they were carrying $775 worth of the work in the process inventory for each of the car they built, but the Japanese carried only $150 of the amount. The only very existence of the Unites states auto industry which depended on the adopting of the JIT philosophy. Very efficient way of transportation methods were one of the critical to the effective JIT process.

Key issues of the transportation before the JIT include plant locations, were the role of the transportation along with the deregulation as well as the transportation regulation. The placement of the plants also changed as a solution of the implementation of the JIT.

Homework Help: What is the average kinetic energy of each helium atom...

Q)The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
What is the average kinetic energy of each helium atom?
Answer in units of J.

Sol:

The average kinetic energy is given by:

KE_atom = (PV)/N

Number of atoms (N)= Avogadro's number (A) * n
where N = A*n = A*(PV/RT)

N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms

KE_atom = (96235*0.01414 /3.456 x 10^23 )J/atom
      = 3.94 X 10^-21 J/atom

Chemistry Homework Help: How many atoms of helium gas are required to fill a balloon to diameter..

Q)The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
How many atoms of helium gas are required to fill a balloon to diameter 30 cm at 12◦C and 0.95 atm?

Sol:

PV = nRT
n = PV/RT
Number of atoms (N)= Avogadro's number (A) * n
N = A*n = A*(PV/RT)
P = 0.95 * 101300 Pa
P = 96235 Pa
V = (4/3) pi R^3
R = 0.15 m
V = (4/3) pi (0.15 m)^3
V = 0.01414 m^3
T = 273.15 + 12
T = 285.15 K

N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms

HomeWork Help: What is the current in the bulb..

Q) A 100 W lightbulb is plugged into a standard 120 V outlet. (a) How much does it
cost per 31-day month to leave the light turned continuously? Assume electrical
energy costs US$0.06/kWh. (b) What is the resistance of the bulb? (c) What is the current in the bulb?

Sol:

First let us find the energy used in 31 days is(in kWh) and then let go for calculating the cost.

a) U = P * t = 0.100kW * 31days * 24hrs/day

       = 74.4kWh

Cost per 31 day month = 74.4kW h * $0.06/kW h

  = $4.464

b) The resistance of the bulb is given by,

P = V²/R

R = V²/P

  = (120V)²/100W

=144Ω

c) The current in the bulb is given by,

P = VI

I = P/V

     = 100W/120V

 = 0.833 A

Physics Homework: find the speed of the block after it has moved?

Q) 19 kg block initially at rest pulled to the right along a horizontal surface by a constant, horizontal force of 16.1 N. the coefficient of kinetic friction is 0.145. the acceleration of gravity is 9.8 m/s^2.find the speed of the block after it has moved 3.35 m.

Sol:

As per the concept that the work done by the force is equal to the Kinetic energy of the block, we have

u*(mg)*s=1/2mv² where work done= u*(mg)*s, u=0.145,m=19kg,g=9.8m/s², s=3.35m

Therefore by substituting the values in the above equation, we get,

0.145*9.8*3.35=0.5* v²

Hence Speed of the block after it has moved 3.35m is,

V= 3.086m/s

Physics Homework: What is the average force exerted on the ball by the wall?

A )3.45 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60 degrees with the surface. It bounces with the same speed and angle. If the ball is in contact with the wall for 0.200s what is the average force exerted on the ball by the wall?

a) X-component? (answer in N)
b) y-component? (answer in N)

Sol:

   a)  M=3kg, U=10m/s,V=-10m/s, angle=60, t=0.2sec

V=U+at

-10=10+a*0.2

a=-20/0.2

 =-100m/s

F=ma=3*(-100)

    =-300N

F_x = Fcos(60)

  = -300cos(60)= -150N

  b) F_y = Fsin(60)

  = -300sin(60)= -259.80N