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Nairaland Forum / Nairaland / General / Education / Nairaland Jamb Tutorial Centre {physics Thread} (11980 Views)
2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } / Nairaland Jamb Tutorial Centre {chemistry Thread} / Nairaland Jamb Tutorial Centre. {NJTC} (2) (3) (4)
Re: Nairaland Jamb Tutorial Centre {physics Thread} by fasodecapo(m): 12:03pm On Dec 08, 2014 |
VEVEDIHNO:Total distance = 14,840m. check picture below
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Re: Nairaland Jamb Tutorial Centre {physics Thread} by VEVEDIHNO: 9:41am On Dec 15, 2014 |
plz u guyz shuld help me with diz An anti air craft shell is fired vertically upward with a muzzle velocity of 1km/s (a)calculate the maximum height it can attain (b)the time taken to reach this height (c)the instantenous velocity at the end of 20s and 50s (d)when will it height be 37.5km tanks in advance |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by Adiwana: 8:33am On Dec 23, 2014 |
VEVEDIHNO: Using the equation of motion ''V^2=U^2+2ah, at max.height V=0 U^2=1km/s(1000x1000m/s),a=-10 since its moving upwards[when moving downwards a is +],h=? 1).0^2=1000^2+2(-10)h 0=1000000-20h 20h=1000000 h=1000000/20 h=50000m h=50000/1000 h=50km 2)v=u-at 0=1000-10t 10t=1000 t=1000/10 t=100s .:.t=100s 3). Inst.velocity v=u+at v=1000-10(20)=1000-200=800m/s v=u+at v=1000-10(50)=1000-500=500m/s 4).h=37.5x1000=37500m s=ut+1/2gt^2 37500=1/210t^2 37500=5t^2 t^2=37500/5 t^2=7500 t=√7500 t=86.6s |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by gameboi: 10:38am On Dec 23, 2014 |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by gameboi: 10:39am On Dec 23, 2014 |
VEVEDIHNO:(a) From v2=u2+2gh U, initial velocity=1km/s=1000m/s V=0m/s, g= -10m/s2( upward vertical motion), h= ? Therefore v2-u2=-2gh Substituting all values, we have 0-(1000)^2= -2(10)h -(1000)^2= -20h H= - (1000000)/ -(20) = 50000m Maximum height=50000m (b) from V=U+gt V-U=gt. Where g still remains -10m/s2 and U=1000m/s 0-1000= -10t T= -(1000)/ -(10). T=100secs time taken to attain maximum height=100s (c) Instantaneous velocity, V=u+gt i. When t=20s, V=1000+(-10)x 20 = 1000-200= 800m/s ii. When t=50s V=1000+(-10)x 50= 1000-500= 500m/s (d) from V2=U2+2gh when H,height= 37.5km or 37500m H= u2/2g 37500= u2/2(10), U2=37500x20=750000 U= root750000= 866.03m/s Time= u/g Time=866.03/ 10= 86.6secs Therefore when the height is 37500m, the velocity would be 866m(approximately) and time would be 86.6secs. To save yourself of all the torture of deriving those formulas, use the following formulas Maximum height, Hmax= U2/2g Time taken, t= U/g Time of flight= 2u/g |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by shakrullah(m): 3:34pm On Jan 14, 2015 |
gameboi:modify ur post. Maximum Height is =U^2/2g not U2 |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by gameboi: 5:14pm On Jan 14, 2015 |
shakrullah:Boss, Seriously ? Apart from that, were the solvings correct? |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by shakrullah(m): 6:40pm On Jan 14, 2015 |
gameboi: 100% correct |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by VEVEDIHNO: 4:11am On Jan 22, 2015 |
Sum1 shuld assist on this question A alloy of mass 588g and a volume 100cm^3 is made of density 8.0g/cm^3 and alluminium of density 2.7g/cm^3. Calculate the proportion; (i)By Volume (ii)By Mass of the constituents of the alloy. Fankz in advance. |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by lisimmo(m): 7:02am On Jan 22, 2015 |
Jamb examination is already here Practice JAMB, POST UTME CBT questions online for free. You can compare your scores with other aspirants and candidate nationwide. You can also monitor your improvement. at www.e-testpedia.com Sample questions below:
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Re: Nairaland Jamb Tutorial Centre {physics Thread} by Nobody: 1:04pm On Jan 27, 2015 |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by Adiwana: 5:38pm On Feb 01, 2015 |
VEVEDIHNO:let mass of Aluminum=X p=m/v v=m/p 100=588-X/8 + X/2.7 100x8x2.7=8X+2.7(588-X) 2160=8X+1587.6-2.7X 2160-1587.6=8X-2.7X 572.4=5.3X X=572.4/5.3 X=108g mass of Al=108g but p=m/v v=108/2.7 v=40cm^3 |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by Vecto(m): 9:47am On Jun 05, 2016 |
Kunlexic: Hello genius..Please can you help with this? Im having difficulty solving it especially 2. A steel is 8m long and 4mm in diameter at 0°c, calculate 1. The length of the wire at 60°c 2 the cross sectional area at 100°c 3 the increase in volume of the wire at 70°c Given that the linear expansivity of the material of the wire is 1.2* 10^–5 K^–1 Thanks. |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by princeivon(m): 7:20am On Feb 10, 2017 |
pls guyz help me out on dis question... A glass bottle full of mercury has mass 5OOg. On heating through 35c of mercury are expelled. Calculate d mass of mercury remaining in d bottle. Cubic expansivity of mercury is 1.8x1O raise to power 4. Linear expansivity of glass is 8.Ox1O raise to power 6 |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by CustomII(m): 12:00pm On Mar 23, 2018 |
my first assignment here is are; ( A stone is trow veritically upwaed from the ground with a velocity of 40m/s calculate) a. the maximum high reached. b. the time taken to attain the maximum height. c. the time to reach the ground. d. the velocity reached half way to the maximum height. the second question; an inclined plane 6meter has one end raise by two meter if the efficiency of this machine is 72% a.what is the minimum effort require to raise the load of 100N of the plane. b. what will be the effort if the plane is perfectly smooth. |
Re: Nairaland Jamb Tutorial Centre {physics Thread} by CustomII(m): 12:01pm On Mar 23, 2018 |
please helps my nairalander gurus CustomII: |
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