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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 OGOJNR(m): 8:51pm On Jun 06, 2015 |
hmmm...try again |
Re: University Of Ibadan 2015/16 Applicants by mathefaro(m): 8:51pm On Jun 06, 2015 |
horpeyemmi66:you didn't consider the fact that the ball has only traveled for 5s, the height Won't be 50m. Anyway, second half I'd about to begin. See you guys later |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 9:09pm On Jun 06, 2015 |
K.E=1/2mv^2 K.E= 1/2m(gt)^2 K.E=1/2 x 65 x (10 x 5)^2 K.E= 81,250J |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 9:22pm On Jun 06, 2015 |
yea,i nid more of ur physics questn@ope |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 9:32pm On Jun 06, 2015 |
Solve the last unsolved question I posted then we can continue from there... |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 9:36pm On Jun 06, 2015 |
got 69.67 bt nt sure |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 6:54am On Jun 07, 2015 |
A BODY SLIDES DOWN AN INCLINED PLANE WHICH MAKES ANGLE OF 30 WITH THE HORIZONTAL,NEGLECTING FRICTION.CALCULATE: a.the velocity of the body after sliding 10m from rest b.the time taken to slide the 10m distance MORNING TO Y'ALL AND HAPPY SUNDAY |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 8:52am On Jun 07, 2015 |
OGOJNR:Good morning to you too... Ans to your questions F - mgsinΘ - Fr = 0 F - mgsinΘ - 0 = 0 F - mgsinΘ = 0 F = mgsinΘ ma = mgsinΘ a = gsinΘ a = 10 x sin 30° a = 5ms^-2 Then Velocity can be found thus, V^2 = U^2 + 2as V^2 = 0 + 2 x 5 x 10 V^2 = 100 V = 10m/s Consequently, S = Ut + 1/2at^2 S = 0 + 1/2at^2 10 = 1/2 x 5 x t^2 20/5 = t^2 t^2 = 4 t = 2s Its an Immaculate Sunday.... Proudly C.A.C cum Baptist. 1 Like |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 11:55am On Jun 07, 2015 |
I see you @Oluwatobiloba... -------------------------------------------------------- *PHYSICS | EXPANSIVITY 1.A triangular plate of height 6cm and base length of 12cm is made of metal of linear expansivity 3 x 10^-5 K^-1 as the plate is heated from 15°C to 70°C, the area of one side of the plate will increase to? 2. A steel plug has a diameter of 5cm at 30°C. At what temperature will it fit exactly into a hole of constant diameter 4.997cm?[Co-efficient of linear expansion of steel is 11 x 10^-6 c^-1] Cc DrHost, OGOJNR, POlymath, Bhuuumhite, mathefaro, medwhite, funkysilver, Pr0ton, Augster, Horlarmancy, becborn, Luukasz,Francistony, AleXis0r, thaotech, atakitisam, UMartins1, darrytoz et al |
Re: University Of Ibadan 2015/16 Applicants by mathefaro(m): 12:42pm On Jun 07, 2015 |
horpeyemmi66:A1 = 0.5*12*6 = 36cm^2 Superficial expansivity, B = 2æ B = 2*3*10^-5 = 6 *10^-5K^-1 ∆@= 70 -15 = 55°C A2=A1(B∆@ + 1) A2 = 36(6*10^-5*55 + 1) A2 = 36(0.0033 + 1) A2 = 36(1.0033) A2 ~= 36.12cm^2 to 2d.p 1 Like |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 2:23pm On Jun 07, 2015 |
mathefaro:Very correct... |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 2:25pm On Jun 07, 2015 |
Am I correct @OGOJNR? |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 2:50pm On Jun 07, 2015 |
b=2a
0.00006=A2-A1/A1¤
AREA=1/2*12*6
=36CM CUBE
¤=70-15
=55CM
0.00006*36*55=A2-36
A2=36+0.1188
a2=36.1188cm sq |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 2:54pm On Jun 07, 2015 |
a=l2-l1/l1*¤
0.000011=5-4.997/5¤
¤=0.003/0.000055
¤=54
bt ¤2-30=54
¤2=54+30
=84' |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 3:01pm On Jun 07, 2015 |
yea bt can u smash dis? A body of mass 50kg moving with a uniform speed of 5m/s comes to rest in 5 sec.calculate its acceleration before coming to rest. |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 3:37pm On Jun 07, 2015 |
OGOJNR:You wanna play a fast one on me, but I know very well that "time is independent of mass", I guess that 50kg is just a figure-head. 1st equation of motion V= U + at 5= 0 + at 5= 0 + 5a a=1ms^-2 |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 3:43pm On Jun 07, 2015 |
finally,i got u...that questn dnt nid any calculation since u're lukin for d acceleration before cumin to rest...wen a body undergoes uniform velocity or constant speed the acceleration is always zero...the ans is 0 |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 3:44pm On Jun 07, 2015 |
hw did i do in dat expansivity questn? |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 3:46pm On Jun 07, 2015 |
OGOJNR:You wanna play a fast one on me, but I know very well that "time is independent of mass", I guess that 50kg is just a figure-head. 1st equation of motion V= U + at 5= 0 + at 5= 0 + 5a a=1ms^-2 This method, I don't totally like, as the body has to travel through a distance S before coming to a halt, which is Illustrated in the second analysis S= Ut + 1/2at^2 S= 0 + 1/2 x v/t x t^2 S= 1/2 x 5 x 5 S= 12.5m Then, V^2 = U^2 + 2as 5^2 = 0 + 2 x a x 12.5 25 = 25a a= 1ms^-2... |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 3:49pm On Jun 07, 2015 |
hw did i do in dat expansivity questn |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 3:52pm On Jun 07, 2015 |
OGOJNR:For what question? you are not serious o, finally I got a question of yours wrong...sure you did got me....aaaRRRGGhhhh! You got that with 36.12 or there about as the answer very correctly. |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 3:55pm On Jun 07, 2015 |
THE 2ND ONE |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 4:18pm On Jun 07, 2015 |
OGOJNR:nope, try again...use this β= ΔA/ [A1(Θ2-Θ1)] β= A2- A1/[A1(Θ2-Θ1)] Cross multiplying, you have A2 - A1 = β [ A1 ( Θ2 - Θ1 ) ] A2 - A1 = β A1 ( Θ2 - Θ1 ) A2 = A1 + β A1 ( Θ2 - Θ1 ) After factorizing A1 out, A2 = A1 [ 1 + β ( Θ2 - Θ1 ) ] Recall that β = 2α, A2= A1 [ 1 + 2α ( Θ2 - Θ 1) ] (πd^2)/4 = (πd^2)/4 [ 1 + 2α ( Θ2 - Θ1) ] π can cancel out up there... Just insert your parameters, be careful though. |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 4:30pm On Jun 07, 2015 |
itz linear expansivity,hw did it become superficial,sir..i dnt undastand |
Re: University Of Ibadan 2015/16 Applicants by nessie17(f): 4:31pm On Jun 07, 2015 |
Hello every 1 av bin luking at this page as a guest and I lyk the spirit here, I mean serious minds, its gooood. Pray we get admitted. Let's keep the fire burning |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 4:45pm On Jun 07, 2015 |
OGOJNR: The relationships between linear, area/superficial and volumic/cubic expansivity. For Area/superficial expansivity, β= 2α for Cubic/ Volumic, γ =3α α= β/2 So γ is also = 3β/2 α= γ/3 β=2γ/3 ...and thanks nessie17, we all appreciated that timely post. |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 4:48pm On Jun 07, 2015 |
i knw bt PLS solve it...stil got 84 |
Re: University Of Ibadan 2015/16 Applicants by mathefaro(m): 6:17pm On Jun 07, 2015 |
OGOJNR:my brother, it's not linear expansivity. It's actually superficial because Its the cross sectional area of the the plug that needs to decrease to fit in into the hole and not the length. So if you calculate well, you'll get a temperature less than the given one in which the radius is 5cm. I hope it's clear now. **Winks** For me, I'm still at a birthday party(owo epo things) lol |
Re: University Of Ibadan 2015/16 Applicants by OGOJNR(m): 6:23pm On Jun 07, 2015 |
no wonder,i sensed manipulation...i dnt see cross sectional area in d questn nw |
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 6:30pm On Jun 07, 2015 |
OGOJNR:Ok, here is the thing, note the following well because once a mistake is made in them, it just sends all your workings crumbling like a pack of dominoes. A2 = 5cm A1 = 4.997cm Θ2 = 30°C Θ1 = ?? From here, A2 - A1 = β A1 ( Θ2 - Θ1 ) A2 = A1 + βA1 ( Θ2 - Θ1 ) A2 = A1 [ 1 + β ( Θ2 - Θ1 ) ] A2 = A1 [ 1 + 2α ( Θ2 - Θ1) ] (πd^2)/4 = (πd^2)/4 [ 1 +( Θ2 - Θ1 ) π cancels out, (5^2)/4= (4. 997)^2/4[ 1 + 2( 11 x 10^ -6) (30 - Θ1)] 25/4 = 24.97/4[1 +[22 x10^-6] (30 - Θ1)] 6.25/6.242 = 1 + (22 x 10^-6) (30 - Θ1) 1.0012 - 1= 22 x 10^-6 (30 - Θ1) 0.0012/ 0.000022 = 30 - Θ1 54.54 - 30 = -Θ1 -Θ1 = 24.54 → Θ1 = - 24.54°C |
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