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Physics Gurus Pls Do Dis / Physics Gurus Show Ur Face Oh. / Physicist/ Physics Gurus. Lets Find Help Here. (2) (3) (4)

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Nairaland Physics Gurus by Nobody: 8:45am On Aug 20, 2013
I noticed that after JAMB the thread for physics jamb outlived its usefulness, so i decided that i should create this thread for the physics gurus in the house, and thosd who have troubling physics problems. ¤¤~**|PHYSICS|**!¤¤
Re: Nairaland Physics Gurus by topsonkay(m): 9:13am On Aug 20, 2013
Calculate the force acting on a current-carrying conductor of length 2m and radius 5m if placed at 30¤ to the field and the current flowing through it is 3A. (permeability of free space=4*10^-7)

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Re: Nairaland Physics Gurus by Nobody: 9:58am On Aug 20, 2013
topsonkay: Calculate the force acting on a current-carrying conductor of length 2m and radius 5m if placed at 30¤ to the field and the current flowing through it is 3A. (permeability of free space=4*10^-7)
first we need to find the magnetic flux density B=uoI/(2pi.r) then replacing the values, we have B= (4pi*10-7*3)/(2pi*5)=1.2*10-7T. Them the formula for force on a current carrying conductor=BILsin© replacing the values then, (1.2*10-7*3*5*sin30)= 9*10-7N
Re: Nairaland Physics Gurus by Nobody: 10:03am On Aug 20, 2013
topsonkay: Calculate the force acting on a current-carrying conductor of length 2m and radius 5m if placed at 30¤ to the field and the current flowing through it is 3A. (permeability of free space=4*10^-7)
UI/2₹r.....4₹*10^-7*3/2₹5=1.2*10^-7T.........Fm=BILsin¤,1.2*10^-7 * 2 * 3=7.2*10^-7Nm
Re: Nairaland Physics Gurus by Nobody: 10:10am On Aug 20, 2013
ikeola6: UI/2₹r.....4₹*10^-7*3/2₹5=1.2*10^-7T.........Fm=BILsin¤,1.2*10^-7 * 2 * 3=7.2*10^-7Nm
i think you made a slight error, since r=5, then 2r=10, but you wrote 25
Re: Nairaland Physics Gurus by Nobody: 11:19am On Aug 20, 2013
honey: i think you made a slight error, since r=5, then 2r=10, but you wrote 25
no d formula is UI/2₹R....I used dis symbol "₹" to rep pie....bt d pie was cancelled out by d constant U...wich is 4₹*10^-7 bt d guy who postd d questn did nt put pie bt i did...
Re: Nairaland Physics Gurus by Nobody: 11:57am On Aug 20, 2013
ikeola6: no d formula is UI/2₹R....I used dis symbol "₹" to rep pie....bt d pie was cancelled out by d constant U...wich is 4₹*10^-7 bt d guy who postd d questn did nt put pie bt i did...
YOUR FMLA IS RIGHT because PI. WILL CANCEL out
Re: Nairaland Physics Gurus by Nobody: 12:13pm On Aug 20, 2013
Yes tank u
Re: Nairaland Physics Gurus by Nobody: 1:11pm On Aug 20, 2013
ikeola6: Yes tank u
YOU ARE THE BOSS , AND WE ARE STILL YOUR BOYS
Re: Nairaland Physics Gurus by Nobody: 2:02pm On Aug 20, 2013
honey: YOU ARE THE BOSS , AND WE ARE STILL YOUR BOYS
abeg oo m a small boy....i just came to learn 4rm u guys ni
Re: Nairaland Physics Gurus by Nobody: 11:18am On Sep 02, 2013
A 40.0 kg carpenter whose leg bones are 4.0cm^2 in area and 50.0 cm long falls through a height of 2.0 m without breaking his leg bones .if the bone can support a stress of 9.0*10^9 N/m^2, what is the young' s mudulus for the bone' s material? ( if needed use g~9.81ms^-2)..
Re: Nairaland Physics Gurus by Nobody: 12:29pm On Sep 02, 2013
benbuks: A 40.0 kg carpenter whose leg bones are 4.0cm^2 in area and 50.0 cm long falls through a height of 2.0 m without breaking his leg bones .if the bone can support a stress of 9.0*10^9 N/m^2, what is the young' s mudulus for the bone' s material? ( if needed use g~9.81ms^-2)..
modulus(Y)= (F/A)*(L/£) , since the carpenter is falling from a height, the bone will undergo compaction instead of expansion, then we can deduce that the workdone from falling from that height equals the workdone in compacting the elasting bone, 1/2F*£=Mgh, since (F/A)= tensile stress = 9x109Nm-2, A=4cm2= 4x10-4m2, then F= 9x109*4x10-4 = 3.6x106N, making £ the subject of fmla in the former equation, £= [2mgh]/F, replacing the values £= (2*40*9.81*2)/(3.6x106) =4.36x10-4, now that we have gotten £ , strain= £/L, =(4.36x10-4)/0.5 = 8.72x10-4, then modulus(Y) = stress/strain =(9x109)/(8.72x10-4)=1.032x1013Nm-. or 1032GPa , i hope i am correct
Re: Nairaland Physics Gurus by Richiez(m): 2:03pm On Sep 02, 2013
Nice thread!
Re: Nairaland Physics Gurus by Nobody: 2:07pm On Sep 02, 2013
Richiez: Nice thread!
thank you sir.
Re: Nairaland Physics Gurus by Richiez(m): 2:08pm On Sep 02, 2013
honey: I noticed that after JAMB the thread for physics jamb outlived its usefulness, so i decided that i should create this thread for the physics gurus in the house, and thosd who have troubling physics problems. ¤¤~**|PHYSICS|**!¤¤
the jamb thread is still useful and should be considered a sister thread because as it stands, utme will be conducted every year... good job though
Re: Nairaland Physics Gurus by Nobody: 2:12pm On Sep 02, 2013
Richiez:
the jamb thread is still useful and should be considered a sister thread because as it stands, utme will be conducted every year... good job though
POINT TAKEN, THANK YOU AGAIN
Re: Nairaland Physics Gurus by Kunlexic(m): 2:39pm On Sep 02, 2013
A note of 1600Hz is sounded on a day when the speed of sound in air is 320ms^-1.What is the wavelenght in air of this note?
Re: Nairaland Physics Gurus by Nobody: 2:46pm On Sep 02, 2013
Kunlexic: A note of 1600Hz is sounded on a day when the speed of sound in air is 320ms^-1.What is the wavelenght in air of this note?
V(speed of sound) = f/\, making wavelenght the subject of fmla /\= V\f= 320/1600= 0.2metres
Re: Nairaland Physics Gurus by Fynestboi: 2:56pm On Sep 02, 2013
honey: I noticed that after JAMB the thread for physics jamb outlived its usefulness, so i decided that i should create this thread for the physics gurus in the house, and thosd who have troubling physics problems. ¤¤~**|PHYSICS|**!¤¤
hmmmm it didnt u can still continue with the thread

https://www.nairaland.com/1209518/physics-chemistry-guru-meet-here#14516995
Re: Nairaland Physics Gurus by Kunlexic(m): 2:56pm On Sep 02, 2013
Click on dis link to join the chemistry gurus.

https://www.nairaland.com/1422385/nairaland-chemistry-gurus

thanks and welcoming everybody and feel free to post ur questions and give answers to a question.
Re: Nairaland Physics Gurus by Fynestboi: 2:57pm On Sep 02, 2013
@honey satisfied?


Now give me three reason why u think a thread that would still be useful for nxt yr jambites/pume had outlived its uses?

Need answer to this..
Re: Nairaland Physics Gurus by Nobody: 10:28pm On Sep 02, 2013
honey: modulus(Y)= (F/A)*(L/£) , since the carpenter is falling from a height, the bone will undergo compaction instead of expansion, then we can deduce that the workdone from falling from that height equals the workdone in compacting the elasting bone, 1/2F*£=Mgh, since (F/A)= tensile stress = 9x109Nm-2, A=4cm2= 4x10-4m2, then F= 9x109*4x10-4 = 3.6x106N, making £ the subject of fmla in the former equation, £= [2mgh]/F, replacing the values £= (2*40*9.81*2)/(3.6x106) =4.36x10-4, now that we have gotten £ , strain= £/L, =(4.36x10-4)/0.5 = 8.72x10-4, then modulus(Y) = stress/strain =(9x109)/(8.72x10-4)=1.032x1013Nm-. or 1032GPa , i hope i am correct
almost....just multiply your answer by 2

u try bro..
Re: Nairaland Physics Gurus by Odilop: 1:45pm On Sep 05, 2013
Am ingrin
Re: Nairaland Physics Gurus by Nobody: 1:49pm On Sep 05, 2013
Odilop: Am ingrin
oya contribute
Re: Nairaland Physics Gurus by Odilop: 6:45pm On Sep 05, 2013
Whats magnetic moment ?
Re: Nairaland Physics Gurus by Busayorafael(m): 12:52pm On Sep 08, 2013
ikeola6: UI/2₹r.....4₹*10^-7*3/2₹5=1.2*10^-7T.........Fm=BILsin¤,1.2*10^-7 * 2 * 3=7.2*10^-7Nm
an observation boss. The answer supposed to be 3.6*10^-7.. You know f=bilsin¤ .... so f=1.2*10^-7*3*2*0.5(sin 30=0.5). Therefore f=3.6*10^-7. Thanks

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Re: Nairaland Physics Gurus by Busayorafael(m): 1:01pm On Sep 08, 2013
Question for The boss in the house... A copper rod has a diameter of 4cm and a length of 0.8m. One end is placed in boiling water and the other end is placed in an ice bath. How much heat is conducted through the bar is 1 minute? (kc=380Wm-1K-1)
Re: Nairaland Physics Gurus by Nobody: 4:36pm On Sep 08, 2013
Busayorafael: Question for The boss in the house... A copper rod has a diameter of 4cm and a length of 0.8m. One end is placed in boiling water and the other end is placed in an ice bath. How much heat is conducted through the bar is 1 minute? (kc=380Wm-1K-1)
heat flow rate is given as dQ/dT= (kc*A*(T2-T1))/L.... finding the parameters, A= 2*pi*r*h= 2*3.142*0.02*0.8= 0.1m2.... T2-T1= temp. of boiling water -temp of ice = 100-00= 100.... L=0.8, subtituting the values in the equation, we have dQ/dT= (380*0.1*100)/0.8= 4750Js-1.... the heat conducted in one minute Q = 4750*60= 285000J/0285KJ

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Re: Nairaland Physics Gurus by Busayorafael(m): 5:11pm On Sep 08, 2013
honey: heat flow rate is given as dQ/dT= (kc*A*(T2-T1))/L.... finding the parameters, A= 2*pi*r*h= 2*3.142*0.02*0.8= 0.1m2.... T2-T1= temp. of boiling water -temp of ice = 100-00= 100.... L=0.8, subtituting the values in the equation, we have dQ/dT= (380*0.1*100)/0.8= 4750Js-1.... the heat conducted in one minute Q = 4750*60= 285000J/0285KJ
Wow!!! You're really good.. But there is a little mistake...
Re: Nairaland Physics Gurus by MrCalculus(m): 11:46am On Sep 09, 2013
plz need ur assistance::
A man facing a high wall,notice that an echo is heard 4.0s after he makes a sharp sound.After walking 200m directly towards the cliff,an echo is heard after 3.4s.Calculate
(1)the speed of sound
(2)how far he was from the walk when he made his second observation of the echo?
Re: Nairaland Physics Gurus by Nobody: 7:44pm On Sep 09, 2013
Mr Calculus: plz need ur assistance::
A man facing a high wall,notice that an echo is heard 4.0s after he makes a sharp sound.After walking 200m directly towards the cliff,an echo is heard after 3.4s.Calculate
(1)the speed of sound
(2)how far he was from the walk when he made his second observation of the echo?
V(velocity of sound)= 2X/T,
i) in the first case, T=4, then 4V=2x, dividing through by 2,
X-2V=0---------------------(i)
ii) case 2, the distance from the wall was reduced by 200, time(T)= 3.4, therefore
(2X-200)=3.4V
2(x-100)=3.4V, dividing through by 2
x-1.7=100----------------------(ii), bringing the two simultaneopus equations together,

X-2V=0
x-1.7=100, subtracting eqn ii from i
v=333.33, x=666.67
a) speed of sound= 333.33ms-1
b)distance from the wall in case 2 = (2x-200)/2= (x-100)= 666.67-100= 566.67m
i hope i am correct.
Re: Nairaland Physics Gurus by udswagz: 8:50pm On Sep 09, 2013
Hello people, this question & answer website for students
www.thestudybox.com
needs beta testers. If you are a student pls try it out,
ask a question or answer one now !

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