Please how can I upload my music on boom player�

Am really grateful, nice work sir.

DrBESTJC:


Very correct.....according to ONE TIME SUCCESS IN JAMB and my physics teacher as well....

Thanks alot.

Let's considered this also. If we are to considered the resistance of the generator and the appliance which I think should not be then it will go thus;
The 20A current supplied by the generator passes through its 1R internal resistance, and that resistance dissipates I^2 * R = 20^2 * 1 = 400 Watts. The voltage drop across that resistance, at 20A load = I * R = 20 * 1 = 20 Volts. hence the generator’s unloaded voltage of 240V drops to 220V at 20A load. If you’re asking the ratio of the power supplied TO THE EXTERNAL LOAD to the power dissipated IN THE GENERATOR’s INTERNAL RESISTANCE, that ratio is 220V * 20A = 4,400W to 400W or 11:1. 1/12 of the power generated is dissipated in its internal resistance and 11/12 is dissipated in the load resistance.
I hope u get this.

Myketuale:
Am new 100lvl biochemistry student. How can I pass calculus?
What are the basics and the prerequisite?
Thanks

The best is to start from the basics
Get a class VI Futhermath textbook, sacrifice a day to study it, then u'll notice it is as simple as even normal arithmetic. Read painstakingly.
U won't get enough required basic here unless you do it yourself.

DrBESTJC:
Still waiting for the solution.......JAMB is hereooooo,help out please....

I can guess that the recent answer to this is not well detail I should say not correct.
One thing I will add before my point is °all necessary parameters are giving in the question already°
So knowing our V(supplied)=240 and V(deposited)=220 and I=20. Note this °it's because of the resistance present in the generator that made the voltage fall.
So from this relation •power=iv•
Here we go power(ratio)= power(supplied)/power(dissipated)
So we can say p=IV(supplied)/IV(dissipated)
°I is common so we got 240/220=12:11

Please don't bamboozle anyone with your invalid fantasy, the list is yet out and not until it is released publicly by the school admin. Management we won't accept any unfamiliar list from any anonymous bloggers.

Pls is there any watsapp group for oou if so add this no.09098805519. Thanks

Please is there any whatsapp group for oou, if so please add up.*09098805519*

ambalie:
Congrats to all the successful candidates who made it through. Congratulations!!!
However, that you did not make it is not a crime, there is always a way out or you wait for next year.
For further inquiries, complain, missing results and other relevant information, please call us on 081-267-393-26 or send a mail to admissionportal.edu@yahoo.com

This almost got me fainted[color=#000000][/color]

suso:
I had 60% mechanical engineering
DE Aspirant

What are my chances

we just have to wait for the cut-off. And if it is in accordance to 2015/2016, it should be 57 so therefore u have hope.

pls what the cut off point for mech engr

HomoSapiien:


From where I copied it from, it's probably genuine. When it'd be coming out is what I am yet to ascertain. Probably early next month.

Besides I can't reply your emails because I don't have the password, you can drop your number so I can add you to the whatsapp group.

Pls add my number on watsapp 08091689131. Wants to study mechanical engineering

zainkay:
help me bro
serial:WRN170932078
pin:827369014548
Exam no:4310252045

please stop giving people your details, you're just wasting the number of your card used

please with 243(jamb score) and math-c4, english-b2, physics-a1, chemistry-b3, f-math-b3
what my hope for mech engineering

*please help a bro, i got 243 1a 3b's 1c. please what my hope in unilorin as a mech. engr
illicit i need your help o.