Physics Assignment Help: 30-kg hanging object is connected by a light..

Q) 30-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as 0.162, find the tension in the string.

Sol:

It’s given that M1=5kg, M2=30kg, Also M1 is moving with acceleration a, caused by the Tension in string: T_s, and the friction, F_r, Hence aM1= T_s - uM1g, where u= coefficient of kinetic friction.

Similarly, the mass 30kg is moving with the same above acceleration a by the forces of weight and the Tension in the string,

Hence aM2=M2g- T_s

By adding the above two equations we get,

a= (M2g-uM1g)/ (M1+M2)

Therefore the tension in the string is given by,

T_s = M1g (M2-uM1)/ (M1+M2) +uM1g

By substituting the values in the above equation, we get

T_s = [5*9.8*(30-0.162*5)/ (5+30)] +0.162*5*9.8

  = 48.8N

Write Ups: Data Security~Institute of Research has sensitive information..

Q) ABC Institute of Research has sensitive information that needs to be protected from its rivals. The Institute has collaborated with XYZ inc. for research on genetics. The information must be kept top secret at any cost. At ABC Institute, the researchers are unsure about the type of key (Asymmetric or Symmetric) to be used. Please formulate a possible solution and describe the advantages and disadvantages of any solution employed.

Sol:

Here the security measures have to be taken care for both ABC institute as well as XYZ Inc. So the secure information has to be maintained and preserved from both the sides.

First it’s better to have a secure room which we can call it as ODC where all the research is being done. This room must have a user entry access and user validation. The access must be given to very few members who are deeply involved in research. This will help to preserve the information at least within the room. Also only the genuine users have been given access to that room, which helps in tracking. In case if we find any problem or issue then we can catch hold of only those genuine members.

Try to block the emails that come in or go out with large attachments. It’s better to completely scan the machines and emails of the members thoroughly for every two days. Also keep track of the emails and machines of others i.e., who is not a genuine member but works for that company into other field. So that in case of any doubts they will be caught in this scanning itself.

All the telephone lines from ODC block i.e., a place where research of genetics is being done and preserved in both companies have to be recorded or tapped. So that in incase of emergency it will be very helpful to block the leakage that is taking place most probably within the genuine members.

 All the USB drives has to be disabled in all the machines in ODC, so that members will not be able to takeout any confidential data from the system and they will not be able to copy it into any other external drive or Pen drives or other USB dongles.

The concept of data masking can be used here which hides the specific data or some sensitive information within a database from all the unauthorized members so that if any others receive an email from ODC machines they will definitely not have authorization to view that information. 

It’s always better to encrypt the data before storing it in a database or hard disk. Whenever you want to continue with the research then you can decrypt the information with the help of the decryption password. The decryption password has to be randomly generated code which will change every minute. The device that can be used here is secure id device, which randomly generates the numbers for every minute. This device has to be maintained in the ODC, which has to be taken care only by two of the senior members. Anyways without decrypting we will not be able to view the information.

The one and only main disadvantage of this solution is that the initial investment for infrastructure including equipments and man-hours is very high, but once implemented then it will be very useful.

References:

1) Data security masking, 2009

2) Data security erasure, 2009

3) Data security recovery, 2008

4) Disk security encryption, 2010

Computer Science Assignment Help: Consensus algorithm

“Consensus algorithm”:  A group of ten people need to decide which one flavor of ice cream they will all order, out of three options. The algorithm can question and re-question the participants, and present the answers to the participants, until a consensus is reached. This exercise is somewhat more open-ended. Add your assumptions if necessary. Obviously, this algorithm might never result in an answer. Deal with that too.

Answer:

Step1: Please ask the group of people to select one flavor of ice cream and go to step 2.

Step2: Please plan for ordering the selected flavor and go to step 3 else go to step 1.

Step3: Please confirm from all the ten people whether they are ok with the selected flavor of ice cream, if yes go to step 4, else go to step 1.

Step4: Please go for the second round of confirmation from each one of them among the 10, if they are ok then go to step 5, else go to step 1.

Step5: Please order the selected flavor of ice-cream as a group of 10, if there is no discrepancy in the choice and go to step 6, else again go to step 1

Step6: Please wait for the waiter to serve the ice cream, if yes go to step 7, else go to step 8.

Step7: Enjoy having the delicious flavor of ice-cream, go to step 9.

Step8: Please remind the waiter again, and go to step 9.

Step9: Come out of the ice-cream shop as a group of 10 once you finish having it, else go to step 7.

Explanation:

Here we will not consider the percentage of members in a group opting for a flavor choice; instead we go for probability considering three to four members at a time. Hence we give three chances for the group of 10 members to select their choice, we are not ordering at the first choice we are giving two more other chances to choose so that as per the probability, at least after the third round for asking for the choice they all group of ten will come to the conclusion and they must be able to order it at that stage or else this keeps repeating until they all come to the conclusion.

Networking Homework Help: Explain why security is important for WLANs, even more ...

Q)Explain why security is important for WLANs, even more so than for wired networks. 

Sol:

More security is needed and security is very much important in case of wireless LANs, when compared to wired LANs. The main reason behind this is in case of wireless LANs by making use of the wireless authentication of the authorized wireless user any one can make use of it also hacking the wireless network is very easy if it is not password protected. Since in most of the WLANs the users are not protected by a unique password, since wireless LANs are always used in case of places like college campus or office campus where multiple users can access the network. Hence in case of wired LAN there is a need for the dedicated channel or cable through which every user can get connected to the network whereas in case of wireless LANs there is no need for it, sometimes it will not even ask for any security passwords. Hence any user can get connected to the network and use it without paying for it. As a result more security is important for WLAN’s when compared to Wired LAN.

Networking Assignment Help: What specific steps can be taken to improve security in WLANs...

Q)What specific steps can be taken to improve security in WLANs?

Sol:

 The below steps can be taken to improve the security in WLANs:

a) Initially wireless network has to be authorized and authenticated.

b) Also all the users try to get connected to that particular has to be authorized and authenticated.

c) Every user connected to the network has to be uniquely identified by their own unique username and password.

d) Multiple users with same username and password should not be allowed to connect to the network.

e) Every user has to be properly disconnected so that no one in queue must be able to continue to get connected to the network with their username and password.

Think and Answer Assignments: Minicase IV Household Wireless page....

Q)Your sister is building a new two story house which measures 50 feet long by 30 feet wide and wants to network her family’s three computers together. She and her husband are both consultants and work out of their home in the evenings and a few days a month.  Each has a separate office with a computer, plus a laptop from the office that they occasionally use. The kids also have a computer in their playroom. They have several options for networking their home:

a) Wire the two offices and playroom with Ethernet cat5e and put in a 100Base-T switch.

b) Install one Wi-Fi access point ($85) and put Wi-Fi cards in the three computers (the laptops already have Wi-Fi)

c) Any combination of these options

There are many possible solutions to this problem.  What would you recommend, considering their current hardware configuration as well as potential expansion? Justify your recommendation.           

Sol:

The best possible solution is to go for installation of Wi-Fi access points followed by installation of Wi-Fi cards for the desktop computers. Anyways laptops will definitely have Wi-Fi inbuilt so there is no need for separate installation of the cards in laptops. The reason behind this is looking at the future definitely the desktops are getting replaced by laptops. Also laptops are portable, which means we can carry them any where we need but if they are connected to some network cables then we will not be able to access the network wherever we carry the laptops. Also in case of desktops if we place it in 1st floor but tomorrow we need it in the last floor then there is need to work on the network connectivity in it only labor is involved in shifting but network connectivity is always there since it is independent of the location within the office campus or two story house limits. Hence the recommendation here is to go for Wi-Fi connection.

Networking Homework Help: There is an ever growing range of physical media being offered such as fiber...

Q)There is an ever growing range of physical media being offered such as fiber, coax, satellite, IP over power lines, a number of wireless technologies, and more. Recently, the telephone companies announced 4G wireless communications and the IEEE has WiMax at Gigabit rates. What criteria would you use to select technologies for your business and for your home?

Sol:

The following are the criteria being used to select technologies for my business:

a) Multiple connections with maximum speed.

b) Higher bandwidth with lesser price.

c) Unlimited downloads with no additional charges.

d) Unlimited access within the coverage area irrespective of the machines being used.

e) Full time high speed networks without any congestion.

The following are the criteria being used to select technologies for my home:

a) Not more than two connections at a time.

b) Very lesser cost, irrespective of download limits.

c) Limited access but reduction in the monthly bill.

d) Moderate speed and bandwidth, irrespective of the congestion.

e) Lesser monthly rentals provided there are additional charges for download if it crosses the available limit.

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.

Read and Answer: The International Centre for Radio Astronomy Research...

1) The International Centre for Radio Astronomy Research

Sol:

The International Centre for Radio Astronomy Research is a collaborative centre placed in Perth, Western Australia. They are already achieving research excellence in both astronomical science and engineering. The International Centre for Radio Astronomy Research has grown steadily from its launch in September 2009. An equal joint venture between The University of Western Australia and Curtin University, International Centre for Radio Astronomy Research has strong support from the West Australian as well as Australian governments and is working very closely with industry and research partners in Australia and in the entire world. Wireless internet (WIFI) inventor honored Curtin University has been recognized as Dr John O’Sullivan’s ground breaking work in astronomy and wireless technologies by bestowing John with an Honorary Science Doctorate. The latest telescope puts focus on high school discoveries where a new internet telescope was launched by State Education Minister Dr Elizabeth Constable at The Western University of Australia. Also
Sky Noise and Sensitivity Studies of Array Configurations and Antenna Elements were done for the Low-Frequency Band of the SKA.

Questions:

a) What is WIFI?

b) Expand ICRAR?

c) Internet telescope was launched by whom?

d) Who invented wireless internet?

Sol:

a) Wireless internet/wireless LAN.

b) International Centre for Radio Astronomy Research.

c) Dr. Elizabeth.

d) Dr John O’Sullivan

Read and Answer: De-noising sea noise data....

1) De-noising sea noise data

This work presents De-noising sea noise data and also how data from a marine seismic survey, which is highly contaminated by noise, was de-noised through the use of a time-frequency filter. In addition to the present standard work flow designed to remove cavitation noise, strumming noise and swell-noise. We introduce a new approach to track seismic interference. This is done by applying time-frequency de-noising on the slowness that gathers. As a motivation and background for the processing we also explain some of the techniques behind the generation of the many different noise types.

The end story is that even in case low if there is a low quality input data then also it can be turned into high quality seismic sections. Even though the dataset used is only one, the results and its methodologies represented here are considered to be general, and should be definitely applicable to a large number of seismic surveys.

Questions:

a) Name the two components which seismic data consists of? 

b) What is the second problem in this dataset?

c) What is the frequency range of swell noise?

d) What is the frequency range of hydrostatic pressure variation noise?

Sol:

a) signal and a noise component

b) Swell-noise

c) (1-10(15)Hz)

d) (0-1(2)Hz)

The total number of partitions of a k-element...

Q) The total number of partitions of a k-element
   set is denoted by B (k) and is called the k-th Bell number.

    Thus B (1) = 1and B (2) = 2.
   Need to find B (k) for k = 4, 5, 6.

Sol:

We know that,

Where,

Is the Binomial Coefficient.

It is given by:

It’s given that B (0) =1; B (1) =1; B (2) =2

Hence for n=3, from the first formulae

B (3) =C (2, K) B (K) where K=0 to n-1

B(3)=C(2,0)B(0)+ C(2,1)B(1)+ C(2,2)B(2)

From the second formulae, C (2, 0) =2! /0! (2-0) = 1

C (2, 1) =2; C (2, 2) =1

Hence,

B (3) =1*1+2*1+1*2=5

For n=4,

B(4)= C(3,0)B(0)+ C(3,1)B(1)+ C(3,2)B(2) + C(3,3)B(3)

        = 1*1+3*1+3*2+1*5= 15

Similarly,

For n=5,

B (5) = 52

For n=6,

B (6) = 203

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

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

Physics Assignment Help: How much work is done moving a...

Q)How much work is done moving a 2.5 kg book to a shelf 3.0 m high?

a.What is the potential energy of the book in #7 above?
b.If the book in #7 above, falls off the shelf, how much kinetic energy will it have when it hits the floor

Sol:

Work done = Force*displacement

Force=m*g=2.5*9.8=24.5N

Work done= 24.5*3=73.5Joules

a) Potential energy = work done= m*g*h= 73.5Joules

b) Kinetic energy when it hits the floor will be at t=0,

V=U + at=0+9.8*0=0

KE= ½*m*v² = 0J

Chemistry Assignment Help: What is the equilibrium concentration of H2S...

Q)Kc = 2.6 × 10^8 at 825 K for the reaction

2H2(g) + S2(g) <===> 2H2S(g)

The equilibrium concentration of H2 is 0.0020 M and that of S2 is 0.0010 M. What is the equilibrium concentration of H2S?

1. 0.0010 M
2. 1.02 M
3. 10 M
4. 0.10 M

Sol:

K_c = [H2S]²/[H2]²*[S2]

2.6*10⁸ = [H2S]²/[0.002]²[0.001]

Hence equilibrium concentration of [H2S]= 1.02M

Write-Ups: You have just been hired as an information security engineer for a large multi-international corporation...

Q) You have just been hired as an information security engineer for a large multi-international corporation. Unfortunately your company has suffered from multiple security breaches that have threaten the public's trust that their confidential data and financial assets are private and secured. Credit card information was compromised by an attacked who infiltrated the network through an vulnerable wireless connection within the organization and the other breach was an inside job, where personal data was stolen because of weak access control policies within the organization which allowed an unauthorized individual access to valuable data. Your job is to develop a risk management policy that addresses the two security breaches and how to mitigate these risks.

This assignment requires 2 to 3 pages in length (500 words minimum), based upon the APA style of writing.

Use transition words, thesis statement, Introduction, Body, Conclusion and Reference Page with at least two references. Double spaced Arial 12 Font.

Need a 2 to 3 page paper on this topic/assignment? Comment on this post!!!

Write-Ups: How do roles simplify database administration with regard to privileges?

Managing privileges can be very confusing if not planned and implemented in an organized manner.
How do roles simplify database administration with regard to privileges?

I would like the answer about 200 words.

Sol:

Managing privileges is not an easy job in an enterprise database. There of thousands of privileges which are being used. Hence Users cannot be assigned and managed with privileges, so we go for something called roles. A set of privileges are assigned to a role. This role is then assigned to users. Each and every user must have a role for accessing any resources in a database.

A role definitely simplifies the work of database administration when compared to privileges. Managing a role is very much easy when compared to managing a privilege. Once the set of privileges are assigned to one particular role the role assigned to the user can be performed within the limits of the assigned privileges.

Let us take an example where the role1 is assigned to the user1 with privileges of insert, update, select and create. Here the user1 has got privileges of inserting the data to a table or updating the data or else just for selecting any table or data or create tables. We can call this as user privileges, since the delete privilege is not given to user. It’s mainly for admin use. Hence user will not be able to delete the data or table but he/she will be able to create a table, update a table or else insert data.

In this manner the set of privileges perform together in the form of a role, which is assigned to users to perform their activity. Hence role simplifies the task of DBA when compared to privileges.

References:

1) Managing Roles and Privileges, 2009

Homework Help: What is the instantaneous power delivered by the...

Q)A single constant force 25.1 N acts on a particle of mass 6.53 kg. The particle starts at rest at t = 0.
What is the instantaneous power delivered  by the force at t = 4.31 s?

Answer in units of W.

Sol:

Force, F=mass(m)*acceleration(a)

a= 25.1/6.53

 = 3.84m/s²

Instantaneous power delivered by the force at time t=4.31sec is given by,

P(t)= F(t)*V(t)

V(t)=U(t)+at

At t=4.31sec, U(t)=0

V(t)=3.84*4.31= 16.55m/s

Hence, P(t)= 25.1*16.55

           = 415.4W

Homework Help: what is the new pressure in the tank?

Gas is confined in a tank at a pressure of 10 atm and a temperature of 13◦C. If 38 percent of the gas is withdrawn and the temperature iGas is confined in a tank at a pressure of 10 atm and a temperature of 13◦C. If 38 percent of the gas is withdrawn and the temperature is raised to 70◦C, what is the new pressure in the tank? Answer in units of atm.s raised to 70◦C, what is the new pressure in the tank?
Answer in units of atm.

Sol:

Here we will make use of the formulae:

P2 = P1*(V1/V2)*(T2/T1), where V2 = 0.62(V1), T1= 13+273= 286K

T2= 70+273= 343

P2 = 10*1.6129*343/286 = 19.34atm