₦airaland Forum

KingErnesto:
Thanks for the attempt, I appreciate. For the question 2, it's actually linear density, kg/m and not kg/s , my mistake.

Velocity = sqrt(4/0.01)=20m/s;

since Sin is 2*pi periodic then the period of the sinusoidal function is 6t=2*pi=> t_p=2*pi/6;
but frequency= 1/t_p= 6/(2*pi)= 3/pi hz;

velocity=frequency*wavelength.

wavelength=velocity/frequency;

wavelength=20/(3/pi)= 20*pi/3 metres.

KingErnesto:
I and my colleagues were solving some physics problems relating to waves. We couldn't arrive at the solution of 2 problems, so we thought perhaps some parameters were omitted. So I suggested that I post a couple of them up here to see if anyone could arrive at a solution....


Questions

1. A steel wire having a diameter of 0.2mm is subject to a tension of 200N . Determine the velocity of propagation if transverse wave along the wire.


2 . A vibrating source at the end of a stretched strung has a displacement given by the equation y=0.1sin6t , where y is in meters and t is in seconds. The tension in the string is 4N and the mass per unit length is 0.01kgm/s .

a. What is the velocity in the strings
b. What is the frequency of the wave
c. what is the wave length.


cc: seunwen2 , privatetisa, simiolu1, STENON, dejt4u, Dominiquez

Question 1 is 900 m/s.

I got this from knowing the density of steel to be 7.86*10^3 kg/m^3. Once you know this then calculate the circular area of the steel wire (which is pi*D^2/4) and then multiply by the density to give you the mass per unit length. It is pretty straight forward from there using the formula V= sqrt(T/M/L).


As for question 2, why is your unit of mass per unit length in kgm/s?. Once you let me know I will comment on question 2.

Fuzzyman:
You should have been assessing only your M.Sc anyway as it is your highest degree.

No it will not affect you negatively when applying for PR

Many thanks for your prompt response.

happytbaby:
You are good to go with just the M.Sc. Ask WES to conclude your evaluation without the B.Sc.

Many thanks for your prompt response.

I added two degrees (B.Sc and M.Sc) in my WES application, I am now thinking of withdrawing my B.Sc due to the ongoing NASU strike. My question is:- is it possible to assess only M.Sc and not B.Sc in WES?. My M.Sc was done in the U.K. while my B.Sc was done in Nigeria; will they be any negative implication when applying for PR?; if the answers to my aforementioned questions are yes and no respectively then how do I go about assessing only my M.Sc given that I have created my WES Profile already and paid for assessment for both degrees?

Mechanics96:
Wrong sir, your answer '3' violates the rules of distribution... 3c4 will never exist. The question is wrong in the first place!

The question is Np3÷NC4=6 and n=7



N! /(n-3)! × (n-4)!4!/n! =6

(n-4)!4!/(n-3)(n-4)! =6

4!/(n-3) =6

6(n-3) =4!

6n-18=24

N=42/6

N=7

It is true that nCr only exist when n>=r >=0 (in other to avoid negative factorial). However, if you narrow down on the meaning of this formula, you quickly realize that it just means how many number of ways in which a given 'r' quantity can be gotten from an 'n' quantity (e.g. 8C5 means how many ways can you choose 5 eggs given 8 eggs). In this scenario 3C4 means how many number of ways can you choose 4 eggs given 3 eggs---the logical answer is 0 ways (because you cannot). So in as much as n>=r for the above stated formula to exist, the converse can indeed be logically derived in the same way 0^0 does not exist (this can be proved) but for convenience some real analysis textbooks assume 0^0=1 (the good ones state this assumption in the footnotes).

PS: n=7 is only correct if the OP admits the question is erroneous and it is the way you have stated it.

Olarewajub:
Some one should please simplify.

2(3)^5-2n

Aybalance:
Also this...If x is a real number and x+11<0 Evaluate /x/divided by x.thanks

-1

Aybalance:
workings pls...

nP3=n!/(n-3)!=n(n-1)(n-2)(n-3)!/(n-3)!=n(n-1)(n-2). -------(1)

nC4=n!/[(n-4)!4!]=n(n-1)(n-2)(n-3)(n-4)!/[(n-4)!4!]=n(n-1)(n-2)(n-3)/4!=n(n-1)(n-2)(n-3)/24. --------(2)

(1)+(2)=6; => n(n-1)(n-2)+n(n-1)(n-2)(n-3)/24=6. ----------(3). Multiply both sides of (3) by 24.

=>24 n(n-1)(n-2)+n(n-1)(n-2)(n-3)=144.----------(4).

=>factorising (4) we have n(n-1)(n-2)(21+n)=144.......(5).

from (5) one can see that n=3 satisfies the equation.


Hence the solution is n=3.

Aybalance:
pls i need help on this... ........nP3+nC4=6.find n

n=3

Red2C:
No. 22 pls

Ok let me have a crack on this.


We were asked to show that 2y^2-y-1<=0 for all values of x in R (i.e. (-inf, inf)), if y=(2x+3)/(x^2+2x+3) ;

Let's first factorize 2y^2-y-1=(2y+1)(y-1);=> (2y+1)(y-1)<=0; for this to be true, it means (2y+1)<=0 and (y-1)>=0 (We will call this scenario 1) OR (2y+1)>=0 and (y-1)<=0 (scenario 2).

Assessing scenario 1(a) 2y+1<=0; => y<=-1/2;(that is y can be -1,-2,-3,... etc). BUT this is in contradiction to scenario 1 (b) y-1>=0 (i.e. -1-1=-2 CANNOT be greater than 0). This means there exist values for y such that scenario one is not satisfied (there goes scenario 1 off the window). We are then left with scenario 2.

We must now show that substituting y=(2x+3)/(x^2+2x+3) into scenario 2 must satisfy the inequality.

Let's substitute it into scenario 2(a), we obtain 2((2x+3)/(x^2+2x+3))+1 ?>=0 (i.e. is this >=0), => 2(2x+3) ?>= -(x^2+2x+3) , simplifying further we have x^2+6x+9 ?>=0; => (x+3)^2 ?>=0. Any number squared is always positive irrespective of whether the number is negative or positive, hence we can now be confident to remove the question mark in front of the inequality and say yes for any value of x, 2y+1>=0, and this implies that y-1<=0 for y=(2x+3)/(x^2+2x+3); hence the proof of 2y^2-y-1<=0 is complete.

Second part (finding the maximum and minimum of y)

y=(2x+3)/(x^2+2x+3); we look for the stationary or turning points which are usually maximum, minimum or inflexion points. The stationary points are points where there is a sign change of y as increases that is y moves from positive to negative or vice versa; this means y must go through 0. Knowing this characteristic, we would want to seek these points ; one way of achieving this is to find the points where the derivative of y wrt x is 0 (i.e. dy/dx=0).

Differentiating dy/dx using quotient rule gives dy/dx=(-2x^2-6x)/(x^2+2x+3)^2=0;

-2x^2-6x=0; =>x^2+3x=> x(x+3)=0=> x=0, x=-3; Hence the turning points occur at x=0 an -3; but you were asked about the maximum and minimum values not where in the x space they occur. Substituting the values into the function we get y=1 at x=0 and y=-1/2 at x=-3.
all other values of y are bounded between 1 and -1/2, which implies the maximum value of y=1 and the minimum value of y=-1/2.

Not reading enough books!

lalla89: Thank you very much with the info I will check the schools even thou september admission has past but wll apply for january,but please like how much will I save in my account before I got my visa at UKBA

post like this justifies justwise ruthless response initially on this thread that i condemned. Pls lalla89 read through this thread before asking questions. I bet you will be glad u did.

Regards

lordbeans

justwise: I make no apology for what i said and will say it again in similar situation. I'm sorry if you don't like it.

A well anticipated response (clap for yourself). But for the sake of posterity i will advise you if you are asked a question that you don't want to answer just ignore it or better still be blunt to tell the person you do not want to answer his/her question(at least a nairalander just confirmed you re very blunt) rather than attacking his person.

Note: I wish not to discuss this any further as i can decipher ur tone from ur response as a person living in his own fantasy where he is king and every other persons re his damn subjects, but in life bro things are not the way they seem, i will however implore you to go back to ur philosophies and dispositions of life i think it will save you a whole world of doom.

justwise: I don't moderate this section for any agency, all possible advice has been given to him, if he can't read and understand this how then can he cope with studying in the UK?

Agency promotion is not allowed here and i'm pretty sure that you know that.

justwise in as much i acknowledge your immense contribution i must say this ur response is generally unacceptable.
please despite this is a faceless forum we should all show respect to one another.