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University Of Ibadan 2015/16 Applicants Call 07037300634 / Federal University Of Petroleum Resources {fupre} 2015/16 Applicants / University Of Ibadan 2014/15 Admission Process.... (2) (3) (4)
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Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 12:31am On Jun 12, 2015 |
OGOJNR:Thanks Sir, this was inspiring. I also wish you all of life's most colourful and God's most beautiful things. |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 1:02am On Jun 12, 2015 |
enry54:As Mathefaro has noted, this is an A-level/Pre-degree question which is a no no for O-levels/PUTME. Why you decided to bring this in I wouldn't know but, nevertheless 1. You use this crazy formula →g(r) = (R/R + H)^2 (acceleration due to gravity of body at a height, not to be mistaken with the gravitational potential of a body at a height V= - GM/ R + H) →g(r)= (6.4 x 10 ^6/ 6.4 x 10^6 + 6.4 x 10^6)^2 →g(r)= 0.25m/s^2 Normally on the earth, g=10ms^-2, so mass can be gotten from there, then W=mg →400N=m x 10 →m=400/10 →m= 40kg therefore, the weight of the Missile at the new Altitude is W= mg(r) W= 40 x 0.25 →W= 10N 2. Am not totally sure about the mass, →W=mg →W= d x V x g →mg= d x v x g →g cancels out, →m= d x v →m= 5500kg/ m3 x m3 →m3 cancels out m3, you are left with →m= 5500kg Applying →g = Gm/R^2 →g= 6.67 x 10^ -11 x 5500/ (6400 x 10^3)^2 →g= 8.95 x 10^-21 m/s^2 3. You get an answer to this later in the morning... |
Re: University Of Ibadan 2015/16 Applicants by augster(m): 1:06am On Jun 12, 2015 |
Guys una no dy sleep ![]() |
Re: University Of Ibadan 2015/16 Applicants by John1305(m): 5:07am On Jun 12, 2015 |
DrHost:Great chances, with prayers... Lots of prayer |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 7:13am On Jun 12, 2015 |
mathefaro:Na Jamb Questions! Based on the syllabus |
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 7:47am On Jun 12, 2015 |
Treasuregr8 view as usual. Cutebumzy where art thee? 1 Like |
Re: University Of Ibadan 2015/16 Applicants by mathefaro(m): 7:48am On Jun 12, 2015 |
enry54:If I hear! I can say it with all authority that jamb will never set such question and neither will any school for PUTME. Because like I said earlier, it's beyond the scope of UTME even if as you put it, it's in the syllabus. For the fact that the topic is part of the syllabus doesn't mean they have to go deep to this level. For instance, when you check the math syllabus, you have calculus in it but there is a limit to which they can go in calculus questions (scope). Nevertheless, I believe horpeyemmi66 has done a good job up there. That should be the correct answer if I'm not mistaken. Good morning to y'all |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 8:33am On Jun 12, 2015 |
mathefaro:. 1• It is a logical question and requires a logical reasoning... The first question is all about the inverse square law of universal gravitation, I believe you know that... Let the mass of earth be M The mass of missile be m While the distance is d Thus. F=(GMm)/d2 Mere looking at the situation, GMm is constant anywhere... Right ? So let F x d2 = constant F1 x d12 = F2 x d22 F2 = F1 (d1)2/ (d2)2 F2 = 400 x (6.4 x 106)2/(12.8 x 106)2 F2 = 400 x (0.5)2 = 400 (0.25) = 100N |
Re: University Of Ibadan 2015/16 Applicants by dara319(f): 8:48am On Jun 12, 2015 |
Hmmmm been awhile here o...tight schedule lately u guyz are really trying here cuz am loving the biology and english here...avnt seen much of chemistry though ...keep up d good work...peepz are benefiting...gudmrng |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 8:50am On Jun 12, 2015 |
mathefaro:2• Let M be the Mass of Earth Let r be radius of the earth Let p be the density of earth g=GM/r2 = G(pV)/r2 V(volume)= 4/3 • πr3 Thus g= Gp • 4/3 • πr g= (6.7x10-11 x 5500 x 4 x π x 6400 x 103)/ 3 Pls cross check with calc. g= 9.87 m/s2 |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 9:08am On Jun 12, 2015 |
mathefaro:3• This, I can say it is sorta advanced but, they are all in our textbooks... For revolution, MV2/r = Mg Mw2r=mg g=(2π/T)2r g=4π2r/T2 T2=4π2r/g T= 2π√(r/g) Hence the mass is not dependent on the period. Note: All questions answered are from O'level related textbooks and not A'level |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 9:53am On Jun 12, 2015 |
enry54:This doesn't go down well with me sir viz: "F2= 400 x (0.5)2= 400 x (0.25) = 100N" F=mg, isn't it?, but look what you've done... F2= W x g F2= mg x g F2= 40 x 10 x 0.25(emphasis mine) You should have gotten the mass as 40 before comparing it with the new gravitation acting on the body at that height. You took the weight of the body on the Earth and multiplied with the gravitation acting on the body at that height, which for the want of a better word is illogical. Mass is constant anywhere, but gravitational force g, varies being less at the equator and more pronounced at poles. Don't confuse people for the love of God! Would soon vet the rest you've done... Cc mathefaro |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 10:11am On Jun 12, 2015 |
horpeyemmi66: The equation you used to find the acceleration due to gravity at a height h above the earth crust is wrong... Check you textbook, let say the acceleration due to gravity above height h is g' with R as the radius of the earth g=GM/R2 Thus g' at height h from the earth crust is g'= GM/(R + h)2 Divide by g Thus, g'/g = R2/(R + h)2 Therefore g' (acceleration due to gravity at height h) = g • R2/(R + h)2 By the way... Check how I got 0.25 I dint find g first... Take a look at the way I solved it once more... |
Re: University Of Ibadan 2015/16 Applicants by thankyouJesus(m): 10:30am On Jun 12, 2015 |
People dey vex for here 2 Likes |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 10:30am On Jun 12, 2015 |
to me ooo...the answer is 10N.using g=GM/R square g=0.25M/S2 W=Mg mass is constant nw w=40*0.25=10N. this is an advanced level question oo |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 10:37am On Jun 12, 2015 |
OGOJNR:Guy! g=GM/R2 Thus g' at height h from the earth crust is g'= GM/(R + h)2 Divide by g Thus, g'/g = R2/(R + h)2 Therefore g' (acceleration due to gravity at height h) = g • R2/(R + h)2 These things are in our textbooks! You don't just consider them |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 10:58am On Jun 12, 2015 |
hen nw,which means g is equal to 0.25 nd mass is 40kg nw |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 11:03am On Jun 12, 2015 |
OGOJNR: g' (acceleration due to gravity at height h ) is not 0.25.... Put g= Which is 10m/s2 R= which is the radius of the earth 6.4x106 h= which is the height above the earth surface 6.4x106 In the equation... g' (acceleration due to gravity at height h) = g • R2/(R + h)2 You will get 2.5 as your answer dude! |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 11:19am On Jun 12, 2015 |
*PHYSICS | LATENT HEAT 001 1. Which of the following lowers the melting point of substances? A. Density B. Pressure C. Energy density D. Specific latent heat of fusion 2. Heat is supplied to a test tube containing 100g of ice at its melting point. The ice melts completely in 1min. What is the power rating of the source of heat(latent heat of fusion of ice=336J/g) 3. The statements below were made by some students describing what happened during the determination of the melting point of solids 1. The temperature of the solid was constant until melting started 2. The temperature of the solid rose until melting started 3. During melting, the temperature was rising 4. During melting, the temperature was constant 5. The temperature continued to rise after all the solid had melted 6. The temperature stopped rising after all the solid had melted. Which of the following gives correct statements in the right order? A. 2, 4 & 5 B. 2, 3 & 6 C. 1, 3 & 6 D. 1, 3 & 5 E. 1, 4 & 5 4. What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100°C to 80°C?( Specific latent heat of steam is 2,260,000J/kg, Specific heat capacity of water is 4200J/kg/k) 5. 80% of the heat supplied to a 30g block of ice at 0°C. Calculate the total heat energy supplied. (Specific latent heat of Ice= 336J/g) 6. Which of the following best explains why a person suffers a more severe burn when his skin is exposed to steam than when boiling water is poured unto his skin? A. Steam is at a higher temperature than boiling water B. Steam possess greater heat energy per unit mass than boiling water C. Steam spreads easily over a wider area of the skin than boiling water D. Steam penetrates more deeply into the skin than boiling water 7. Calculate the time to heat 2kg of water from 50°C to 100°C in an electric kettle taking 5A from a 210V supply (S.H.C. of water =4200J/kg/k) 8. How many grams of water, 17°C must be added to a 42g of Ice at 0°C to melt the ice completely? (S.H.C of Water=4200J/kg/k, Specific latent heat of fusion of Ice = 3.4 x 10^5 J/kg/k). ------------------------------------------------ Humanities, where are you guys, this is not a game of monopoly, we need you guys to do your thing a symphony is what we want. |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 12:08pm On Jun 12, 2015 |
horpeyemmi66:1• B 2• Heat Supplied = Heat Required For Melting Ice H = mL = 100 x 336 = 33600J Power Rating = Energy Used Up/ Time Power Rating = 33600/60 = 560J/s or 560w 3• The correct option is A as temp. Rises before and after melting... Temp. Remains constant during melting... 4• Cooling steam from 100oC to 80oC Comprises of (i) 100oC of steam (gas) - 100oC of water i.e change of state (ii) 100oC of water - 80oC of water i.e change of temperature hence, Heat given out = mL + mcθ Cooling water from 100oC to 80oC, heat given out is mcθ Difference = mL + mcθ - mcθ =mL = 4 x 2260000 =9040000J 5• Quant. Of Heat = 30 x 336 = 10080J Thus. 80% of Total Heat = 10080J Total Heat = (10080 x 100)/80 = 12600 6• (B) Steam possesses greater heat energy per Unit Mass. 7• Quant. Of Heat = mc(θ2 - θ1) = 2 x 4200 x (50) = 420kJ Thus. Quant of Heat = IVt t= 420000/(5x210) t= 400s 8•Mass of Water (kg) 0.042kg Heat Required For Melting = Heat Lost By Water 0.042 x 3.4 x 105 = m x 4200 x (17-0) 71400m = 14280 m=0.2kg or 200g |
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 12:24pm On Jun 12, 2015 |
@horpeyemmi66.. I thought you were about to prove me wrong in my previous question.... I hope you now understand the way I did it? Pls note that I will neva test you guys with a question I don't understand! In case I have a question I might not understand, I will indicate it in such question. |
Re: University Of Ibadan 2015/16 Applicants by jolly94: 12:41pm On Jun 12, 2015 |
horpeyemmi66:1.b 2. P=560w 3.a 4. H=9.03*10^6 5. H=8,064J 6.b 7. t=400s 8. M=200g 1 Like |
Re: University Of Ibadan 2015/16 Applicants by thaotech: 3:59pm On Jun 12, 2015 |
enry54: They are actually in the O-levels textbooks. Nice one |
Re: University Of Ibadan 2015/16 Applicants by Umartins1(m): 4:15pm On Jun 12, 2015 |
Wait,,,, Na only scientists full here? No Maths! No Economics! No government! |
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 5:38pm On Jun 12, 2015 |
Umartins1:I can always do Economics, my problem is who will put up Government questions? |
Re: University Of Ibadan 2015/16 Applicants by Umartins1(m): 6:42pm On Jun 12, 2015 |
cassyrooy: I can but only if there is someone to fire me too. |
Re: University Of Ibadan 2015/16 Applicants by Cutebumzy(f): 6:45pm On Jun 12, 2015 |
cassyrooy:hi dear, av been off nairaland for a while. How u doin? |
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 7:37pm On Jun 12, 2015 |
Umartins1:Lets do it then! |
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 7:40pm On Jun 12, 2015 |
Cutebumzy:Welcome back sweetie, how has it been? Hope no stress ni? |
Re: University Of Ibadan 2015/16 Applicants by DrHost: 8:05pm On Jun 12, 2015 |
missing all, where is estijaz ? |
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