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3555 -
modify delete 4128 - Reply from BM (MOROCCO) - 2005-11-26

Hi,

How are you?
I was at China I like it very much.
I hope one day, I can go back.

Have nice day,

3555 -
modify delete 3593 - Reply from wenliwang , 19 yrs (china) - 2005-10-12

hello!I am chinese student.I am 19.I am looking for some foreign pen-friend .I would like to change our study experience and nation culture with you.I think i will make good friend with you .now i introduce myself simply.I am waiting for your letter soon.thanks!

modify delete 3538 - from faycal , 16 yrs (morrocco) - 2005-10-08
Physics - "je veux d'aide pour resoudre un exercice"

salut j'ai un exercices tres difficile en physisque et je cherche quelqu'un qui peut m'aider
un exercice dans le sujet de energie potentielle de pesanteur_energie mecanique
aidez moi svp


3538 -
modify delete 3670 - Reply from Mejbar (FRANCE) - 2005-10-19

TU POURRAIS M'ENVOYER LE SUJET PAR EMAIL
mon email est : bend_100@caramail.com

modify delete 3535 - from lulu (france) - 2005-10-07
Physics - "physique-chimie"

qu'est-ce-qu'une énergie ?


modify delete 3490 - from delphine (france) - 2005-10-03
Physics - "equation aux dimensions"

j'espère que ce coup ci sera le bon!


Bonjour a tous
j'ai un petit probleme avec la correction d'un exercice sur lequel on a passe tres peu de temps.
Je n'arrive pas à retrouver le resultat donne par le prof.Est ce que vous pourriez me dire dans cet exercice comment on arrive a
M-1.L-3.T4.I2
&0 est en fait epsilon de 0



La force qui s'exerce entre deux charges electriques q et q' situees a la distance r l'une de l'autre a pour expression : F=1/(4pi&0)*qq'/r^2

ou &0 est la permittivite du vide.

Quelle est la dimension de &0?




merci d'avance


modify delete 3489 - from delphine (france) - 2005-10-03
Physics - "equation aux dimensions"

je le reposte car il y a des symboles bizarres dans l'autre lol

Bonjour a tous
j'ai un petit problème avec la correction d'un exercice sur lequel on a passé tres peu de temps.
Je n'arrive pas à retrouver le resultat donne par le prof.Est ce que vous pourriez me dire dans cet exercice comment on arrive a
M-1.L-3.T4.I2
&0 est en fait epsilon de 0



La force qui s'exerce entre deux charges électriques q et q' situées à la distance r l'une de l'autre a pour expression : F=1/(4pi&0)*qq'/r²

où &0 est la permittivité du vide.

Quelle est la dimension de &0?




merci d'avance


modify delete 3488 - from delphine (france) - 2005-10-03
Physics - "équation aux dimensions"

Bonjour a tous
j'ai un petit problème avec la correction d'un exercice sur lequel on a passé très peu de temps.
Je n'arrive pas à retrouver le résultat donné par le prof.Est ce que vous pourriez me dire dans cet exercice comment on arrive a
M-1.L-3.T4.I2
&0 est en fait epsilon de 0



La force qui s'exerce entre deux charges électriques q et q' situées à la distance r l'une de l'autre a pour expression : F=1/(4pi&0)*qq'/r²

où &0 est la permittivité du vide.

Quelle est la dimension de &0?




merci d'avance


modify delete 3481 - from iswal (FRANCE) - 2005-10-02
Physics - "Constantes d'équilibre"

Bonjour à tous,

voilà, j'ai un exercice à faire et là je bloque... sad smiley

Voici l'énoncé :
________________________________________________________

Voici les couples suivants : H2S/HS- = 7,0 ; HS-/S²- = 12,9 ; CH3 C02H/CH3COO- = 4,8 ; HF/F- = 3,2

Ecrire la réaction acido-basique et préciser sa constante d'équilibre pour les mélanges suivants :

10mL NaOH (0,10 mol/L) + 10mL (0,10 mol/L)

10mL NaHS (0,10 mol/L) + 10mL HCl (0,10 mol/L)

10mL H2S (0,10 mol/L) + 10mL CH3COONa (0,10 mol/L)

________________________________________________________

en fait mon problème est que je n'arrive pas à établir la réaction acido-basique, ni à calculer les constantes d'équilibre.. sad smiley

Quelqu'un est capable de me mettre sur la piste.. ?? winking smiley

Merci à tous ! winking smiley



3481 -
modify delete 3482 - Reply from iswal (FRANCE) - 2005-10-02

Bonjour à tous,

voila, j'ai un exercice à faire et la je bloque...
Voici l'enonce :



Voici les couples suivants : H2S/HS- = 7,0 ; HS-/S²- = 12,9 ; CH3 C02H/CH3COO- = 4,8 ; HF/F- = 3,2

Ecrire la reaction acido-basique et preciser sa constante d'equilibre pour les melanges suivants :

10mL NaOH (0,10 mol/L) + 10mL (0,10 mol/L)

10mL NaHS (0,10 mol/L) + 10mL HCl (0,10 mol/L)

10mL H2S (0,10 mol/L) + 10mL CH3COONa (0,10 mol/L)





en fait mon problème est que je n'arrive pas à etablir la reaction acido-basique, ni à calculer les constantes d'equilibre.. sad smiley

Quelqu'un est capable de me mettre sur la piste.. ??


Merci à tous !

modify delete 3473 - from Ali (France) - 2005-10-01
Physics - "chimie, niveau 1ere S, electronegativite..."

bon, c'est peut etre une question trop simple mais je suis perplexe. on vient de commencer le cours sur les solutions electrolytiques, et on a aborde l'electronegativite. j'ai compris le concept, mais comment fait on pour savoir si une molecule est dipolaire ou non? ok, je sais que le barycentre des charges negatives ne doit pas coincider avec celui des charges postives, mais comment fait on a determiner l'emplacement du barycentre?? c simple lorsque la molecule est constituee de 2 atomes telle que l'HCl.. mais si on a du NH3 par ex, ou se trouve le barycentre des charges positives?! aidez moi svp!! merci d'avance!


3473 -
modify delete 3487 - Reply from Diana (Belarus) - 2005-10-03

Well, u are plunging into deep water... Do u know what vectors are? Then u can draw the molecule, showing the polar bonds as vectors pointing from positively charged atom to negatively charged atom and check if the vector sum of these vectors is =0 (e.g. methane - weekly polar bonds, but nonpolar molecule) or not =0 (e.g NH3, H2O). Sorry for writing in English but i couldn't express it in French. Get sum1 2 translate it :-) Good luck!

modify delete 3462 - from Jackie (US) - 2005-09-30
Physics - "Relative Velocity"

An aircraft heads due south with a speed relative to the air of 44 m/s. its resultant speed over the ground is 47 m/s The wind blows from the west. a what is the speed of the wind b what is the direction of the aircrafts path over the ground
i figured out a is 16.5 m/s but whats b?


modify delete 3460 - from Emiko (Japan) - 2005-09-30
Physics - "A real tricky one :)"

A skater maintains an average position on the surface of a stream by darting upwards, then drifting downwards to its original position. The current of the stream is 0.500 m/s relative to the shore, and the skater darts upstream 0.560 m in 0.800 s during the first part of its motion. Take upstream as positive direction.

(a) Determine the skater's velocity relative to the water
(1) During its dash upstream
(2) During its drift downstream

(b) How far upstream relative to the water does the skater move during one cycle of its motion?

(c) What is the average velocity of the skater relative to the water?


modify delete 3432 - from Kathy , 20 yrs (France) - 2005-09-27
Physics - "question"

bonjour,
est ce quelqun pourrait me dire qui sont les produits qu'on obtient comme resultat de reaction entre l'oxalate hydraté de fer avec l'hydroxyde de sodium
FeC2O4*2H2O+NaOH


3432 -
modify delete 3486 - Reply from Diana , 22 yrs (Belarus) - 2005-10-03

Je crois, Fe(C204)*H20 + NaOH donneront Na2C2O4 + Fe(OH)2. Fe(OH)2 est puis oxide par l'oxygen de l'air et forme Fe(OH)3. Pardon pour les fautes en Francais. Bonne chance!

modify delete 3421 - from séverine , 19 yrs (france) - 2005-09-26
Physics - "recherche cours"

y aurait il kkun pour m'aider en 2 eme année de licence phisik chimi je galere un peu merci


3421 -
modify delete 4115 - Reply from Yann , 24 yrs (France) - 2005-11-24

Je peux peut-être te filer un coup de main,
je te propose de me contacter par mail ou par messenger sur mon adresse : yannisab@hotmail.fr
bonne continuation dans ta voie ! j'attends de tes news !

3421 -
modify delete 3426 - Reply from anas , 18 yrs (maroc) - 2005-09-26

www.chimix.com contact moi besoin de toi pour m'aider chiki_first@msn.com

modify delete 3362 - from Matt , 18 yrs (Canada) - 2005-09-21
Physics - "Help With Problem!"

Each intersection has a traffic signal and the speed limit it 50km/h. Suppose you are driving from west at speed limit. When you are 10m from the first intersection all lights turn green. The lights are green for 13 seconds each. (a) calculate time needed to reach third stop light. Can you make it through all three lights without stopping?
(b) Another car was stopped at first light when all the lights turned green. It can accelerate at the rate of 2.0m/s^2 to the speed limit. Can the second car make it through all three lights without stopping?

From first light to the second light the distance is 55m. from the second light to third light distance is 85m.


3362 -
modify delete 3390 - Reply from Friend (India) - 2005-09-23

The only trick in this problem is to solve the problem in seperate parts. t1=time taken by first car to reach light 1; t2= time taken from light1 to light2, t3=time taken from light 2-light2

t1=distance/speed=10/(50*(1000/3600))=0.72s
t2=55/(50*(1000/3600))=3.96s
t3=55/(50*(1000/3600))=6.12s

Please note that t1,t2 and t3 each are less than 13 sec so the car would not need to stop at any of the lights and total time will be t1+t2+t3

so second car:

distance(d)=1/2(at^2)
55=1/2(2 t ^2)
t=7.42 sec, so this car will pass light 2 without having to stop as time taken to reach light 2 is less than 13 s. Now the speed of car at light 2 will be v, where v^2=2 a d.

this v will be the initial velocity for going from light 2 to light3, use this to find time taken to reach from light 2 to light 3 and find out if time taken is less than 13 sec or not.

Hope you get the idea. Enjoy!

modify delete 3345 - from Alex (usa) - 2005-09-19
Physics - "help with Physics"

Hey guys...A 52.5-kg high-school student hangs from an overhead bar with both hands. What is the tension in each arm if the bar is gripped with both arms raised vertically overhead?
What is the tension in each arm when the arms make an angle of 40.00 o with respect to the vertical?


3345 -
modify delete 3391 - Reply from Friend (India) - 2005-09-23

It would be best if you make a sketch of the problem, it will be easy to visualize.

there are two arms each having a tension lets say T.

therefore 2T= weight of person (please note that weight =mass X g)

so T=1/2(52.5 X 9.8)

in the second case,

2T cos 45=52.5 X 9.8

modify delete 3344 - from Jordan (usa) - 2005-09-19
Physics - "help with problems for physics"

okaya 130lb wearing spike heel shoes with heals measuring 1/2 " x 1/2" is standing on a linoleum floor calculate how much pressure will be applied to the linoleum if she puts all her weight on One heel?

I know that P=F/A what i get confused on is the 1/2 inch x 1/2 inch do i have to convert those into foot pounds i dont know


modify delete 3308 - from Shaun , 18 yrs (usa) - 2005-09-15
Physics - "perhaps one of you might be able to suceed where I have failed."

O.k. I just spent the last hour of physics 101 trying to solve this problem until I finally got so frustrated that I got up and walked out.

A 2.00m tall basketball player wants to make a basket from a distance of 10.0m. If he shoots the ball at a 45 degree angle, at what initial speed must he throw the ball so that it goes through the hoop without striking the backboard?

I'm sure that this probably very simple and I'm going to feel like a idiot for not getting it but feeling like an idiot is better than not understanding I suppose.


modify delete 3287 - from dani (usa) - 2005-09-11
Physics - "physics 1"

1.an asteroid is heading toward the earth at initial speed 800m/s. someone blows it up. the explosion causes the asteroid to break into 2 parts,smallest part with mass m is pushed forward at a relative speed of 100m/s to the other in the same direction as the initial velocity. the larger chunk is 9 times as massive as the smaller chunk. what is the velocity of the large chunk and small chunk right after the explosion?


3287 -
modify delete 3392 - Reply from Friend (India) - 2005-09-23

lets say mass of smaller chunk is M, then that of larger is 9M. Thus the mass of original asteroid= M+9M=10M

Using conservation of momentum (mass X velocity = constant)

(10M)X 800= M X Vs +(9M) X Vl [Vs and Vm are velocity of small and large chunks respectively]

this gives, 8000=Vs + 9 Vl...........................equ 1

then we use conservation of energy:

1/2[(10M) X 800^2]=1/2[M X Vs^2]+1/2[(9M) X Vl^2]
64 X 10^4=Vs^2+Vl^2 ................................equ2.

Two unknowns, two equations, all yours!

modify delete 3286 - from dani (america) - 2005-09-11
Physics - "physics 1"

in an elastic collision a 6kg mass moving right at 12m/s on a frictionless surface interacts with an 8kg mass moving left 7m/s. using the conservation laws for linear momentum and kinetic energy , compute the final velocities of the masses. check answer by verifying the coefficient of restitution for the collision is 1.

please explain steps and answer to this problem by end of day.


modify delete 3282 - from Megan (USA) - 2005-09-11
Physics - "ahhh! physics help!!"

Can anyone help me solve this problem??

A car with an initial speed of 23.6 km/h accelerates at a uniform rate of 0.98 m/s(squared) for 4.2 seconds.

Find the final speed of the car. Answer in units of m/s.

Find the displacement of the car after that time. Answer in units of km.


modify delete 3281 - from Megan (USA) - 2005-09-11
Physics - "ahhh! physics help!!"

Can anyone help me solve this problem??

A car with an initial speed of 23.6 km/h accelerates at a uniform rate of 0.98 m/s(squared) for 4.2 seconds.

Find the final speed of the car. Answer in units of m/s.

Find the siplacement of the car after that time. Answer in units of km.


3281 -
modify delete 3389 - Reply from Friend (India) - 2005-09-23

1. you have to use the formulae:
v=u+2at,
make sure that the units of all variables is coherent before you get to use the equation. Meaning if u is m/s make sure a is in m/s^2. Since you need the final speed (v) in m/s it is necessary to convert the intial speed(u) from km/h to m/s. So 23.6 Km/H = 23.6*(1000/3600) m/s. Not put all the values and you have 'v'.

2.to find displacement you can either use: d=ut+1/2(at^2) or v^2=u^2+2ad....you will get the answer in meters...to get in kilometeres divide by 1000.

modify delete 3254 - from zhouxz , 25 yrs (China) - 2005-09-09
Physics - "Could you help me?"

Hi everyone!

Who can help me to do the correction of my thesis? I'm a Chinese and I'm not sure if there are any mistakes about gramma in mt english. Thank you and would you send them to my eamil? my address: zhouxz1026@126.com Thank you again:)
Everything except the Integrity
Concerning the Integrity in E-business
Zhou xiaozhi
(03 foreign trade, Zhejiang, Taizhou College)

Abstract:
It is five years since the establishment of e-business. The e-business is at the key point to develop itself into a better one. Although the on-line-shopping conditions have been improved, the problems concerning the safety and honest hesitated customers from take further actions. It is one of the key points to the development of e-business whether we can handle this issue. But tempted by the benefit of dishonest, especially for the opacity of the Internet, the two may possibly don¡¯t know each other very well. Fraud can happen anywhere. There are ways to solve this issue, such as the help from law and education of integrity to Internet users.
Key words:
e-business, integrity, self-discipline in industry.



3254 -
modify delete 3357 - Reply from lulu (france) - 2005-09-21

do you know to do an thesis of physics? I can help you and I had do same lessons of of this subjects .an exemple :the society on the world , the bussiness in the world ...

modify delete 3247 - from joe , 19 yrs (usa) - 2005-09-08
Physics - "help phy2004"

1. (1 point(s))
F1=(0.2,3.7) and F2=(5.8,-6.7) where all components are in newtons. What is the y-component of F1+F2 (in newtons)?




2. (1 point(s))
F=(-3.1,6.4) where all components are in newtons. What is the magnitude of F (in newtons)?






3. (1 point(s))
F=(-5.2,2.3) where all components are in newtons. What angle (in degrees) does F make with the positive x-axis?






4. (1 point(s))
F1=(4.2,1.3) and F2=(-1.5,1.4) where all components are in newtons. What is the magnitude of F1+F2 (in newtons)?




5. (1 point(s))
F1=(0.1,1.3) and F2=(2.4,4.9) where all components are in newtons. What angle does the vector F1+F2 make with the positive x-axis? Express your angle in the range from 0 to 360 degrees.






3247 -
modify delete 3394 - Reply from Friend (India) - 2005-09-23

1.(F1+F2)y=F1y+F2y= 3.7-6.7 = -3
2. F= sqrt[(-3.1)^2+(6.4)^2]
3. F cos theta= -5.2,
F sin theta= 2.3
please note that vector F lies in second quadrent since x component is negative (plot points of F in graph to check)

so tan theta=2.3/5.2

so angle with positve y=axis=90-theta
and angle with poistive x-axis=90+(90-theta)

4. F1=sqrt[4.2^2 + 1.3^2]
F2=sqrt[1.5^2 + 1.4^2]
thus now you can add the 2 to get F1+F2

5. x component of F1+F2= x-component of F1+ x-component of F2
y component of F1+F2= y-component of F1+ y-component of F2
Tan(theta)=y/x

3247 -
modify delete 3393 - Reply from Friend (India) - 2005-09-23

1.(F1+F2)y=F1y+F2y= 3.7-6.7 = -3
2. F= sqrt[(-3.1)^2+(6.4)^2]
3. F cos theta= -5.2,
F sin theta= 2.3

modify delete 3192 - from Rod , 27 yrs (USA) - 2005-08-23
Physics - "Physics Homework"

6. A force of 1.0 x 10^3 N is applied to a steel rod of length 1 m and a circular cross-section of area 10 cm^2. Find the elongation of the rod. Young's modulus for steel is 2.0 x 10^11 Pa.

7. A 300 g baseball traveling at 30 m/s is struck by a bat in the direction from which it came at 40 m/s. Find the impulse of the force on the baseball.

8. An object with a mass of 2 kg traveling with a velocity of 3 m/s, collides head-on and elastically with a stationary object of mass 0.5 kg. The velocity of the 2 kg mass after the collision is:

9. A disk with a mass of 50 kg and radius 0.40 m is rotated about its axis by a motor. The disk starts from rest and rotates at 120 rev/min in 10 s. Find the torque, assuming it to be a constant.

10. A 3 kg bowling ball traveling with a velocity of 3 m/s collides head-on with another bowling ball of the same mass traveling with a velocity of - 2 m/s. If the collision is perfectly elastic, the velocities of the bowling balls after the collision is:


11. A 200 g ball hits a wall perpendicularly with a velocity of 30 m/s and rebounds back perpendicularly with a velocity of 20 m/s. If the collision lasts for 5 milliseconds, find the average force exerted by the ball on the wall.

12. If a solid cylinder, a hoop and a solid sphere of the same mass are rolled down simultaneously from the same height on an inclined plane, in which order will the objects reach the bottom of the incline?

13. A solid cylinder of mass 8 kg and diameter 20 cm is rotated about an axis parallel to its central axis at a distance of 10 cm from the cylinder. The rotational inertia of the cylinder about the parallel axis is:

14. A uniform solid sphere of mass 5 kg and radius 30 cm is rotating about an axis through its center at the rate of 180 rev/min. Find the kinetic energy of the sphere. The rotational inertia of a uniform sphere is 2/5 MR^2.

15. An angular acceleration of 1200 rev/min2 when expressed in radians/s^2 is:

16. A star rotates about its axis through a diameter once in 50 days. At the end of its life, the star collapses to one-hundredth of its original diameter. The rate of rotation of the collapsed star will be:


modify delete 3191 - from Rod , 27 yrs (USA) - 2005-08-23
Physics - "Physics Homework Help"

1. A log of wood is floating in water with 40% of its volume above the surface of the water. If the volume of the log is 0.4 m^3, find the mass of the log.

2. The magnitude of the momentum of a particle is 32 kg m/s at a certain instant. If the velocity has a magnitude of 20 m/s at that instant, the mass of the particle is

3.Water is pumped through a pipe to a tank at the top of a building. The pressure of the water at ground level is 2.2 x 10^5 Pa and at the tank, the pressure is atmospheric. If the diameter of the pipe is the same throughout, find the height at which the tank is located. Atmospheric pressure may be assumed as 1.0 x 10^5 Pa and g as 9.8 m/s^2.

4.The crankshaft of an engine is accelerated from rest at the uniform rate of 4.0 rad/s^2 for 30 s. Find the rotational speed of the crankshaft at the end of 30 s in rev/min.

5. The center of mass of three objects is located at (1, 0). One object with a mass of 5.0 kg is at (- 2, - 1) and a second object with a mass of 2.0 kg is at (0, 0). Find the coordinates of the third mass of 3.0 kg.



modify delete 3173 - from anas , 16 yrs (maroc) - 2005-08-21
Physics - "je cherche des exercices en maths de 6 eme"

je je cherche des exercices en maths de 6 eme merci


modify delete 2897 - from Marta , 15 yrs (Catalonia) - 2005-06-19
Physics - "Help"

Hello...I'm having my final Chemistry test on Tuesday.. could anyone help me? I think I will never understand this subject :(


2897 -
modify delete 3276 - Reply from elena , 13 yrs (usa) - 2005-09-10

try going on a search engine and look up chemistry

modify delete 2817 - from jonas (canada) - 2005-05-28
Physics - "prix de un kwh en russie"

est-ce que quelqu'un pourrait m'aider dans un travail de science
je voudrais savoir quelle est le prix de un kilo watt heure en russie
( kwh ) je serai tres reconnaissant si quelqu'un pouvaitr me le dire
merci beaucoup et bonsoir


2817 -
modify delete 3051 - Reply from Lana (Russia) - 2005-07-27

Bonjour!
Le prix de kw-h en Russie varie (0,84 - 1,1 roubles).
Cela me donnera du plaisir si j' ai aide!

modify delete 2659 - from nadou , 15 yrs (france) - 2005-04-21
Physics - "formule"

qui peux m'aider en physique


2659 -
modify delete 4109 - Reply from FIFI , 16 yrs (ALGERIE) - 2005-11-23

BONJOUR MR,je suis une lyceenne en algerie et j'ai un probleme en phisique mais je suis tres forte en fracais si vous voulez bien repondez sur moi sur : PSSSTOHH@YAHOO.FR

2659 -
modify delete 3824 - Reply from Marcelo , 25 yrs (Brasil) - 2005-10-31

Je m'appelle Marcelo e je suis un eleve de George Washington University en Washington DC USA. Si, vous m'aide avec mon francias je t'aide avec votre physic

2659 -
modify delete 3379 - Reply from sarah , 15 yrs (casa) - 2005-09-22

qu'est ce que tu veux je peux t'aider

2659 -
modify delete 3346 - Reply from Alex (USA) - 2005-09-19

BOJOUR!

2659 -
modify delete 2975 - Reply from console , 18 yrs (congo) - 2005-07-09

j'aimerais bien t'aider mais en condition de me donner tout ton identiter.

2659 -
modify delete 2974 - Reply from console , 18 yrs (congo) - 2005-07-09

j'aimerais bien t'aider mais en condition de me donner tout ton identiter.

2659 -
modify delete 2678 - Reply from rochdi , 17 yrs (marroc) - 2005-04-25

je veux une petite rèsumè du le billan cinètique

2659 -
modify delete 2677 - Reply from rochdi , 17 yrs (marroc) - 2005-04-25

je veux une petite rèsumè du lebillan cinètique

modify delete 2658 - from sonia , 17 yrs (france) - 2005-04-21
Physics - "aide en physique chimie"

bonjour à tous

voilà j'ai un énorme problème j'ai pleins d'exos en physique chimie et je suis nulle en cette matière pourtant j'ai essayé de les faire je n'y arrive pas aidez moi


2658 -
modify delete 4146 - Reply from dounia , 14 yrs (maroc) - 2005-11-27

je veux que vous m'aidez a comprendre la leçon de l'image formeé par une lentille convergente

2658 -
modify delete 3422 - Reply from severine , 19 yrs (france) - 2005-09-26

je peux t'aider di moi ce ke tu nariv pas

modify delete 2656 - from mohamed , 16 yrs (france) - 2005-04-20
Physics - "demande d aide aux question"

Pouvez vous m'aider a repondre a ses questions svp

-comment son place les elements appartenant a une triade?
-pourqoi la classification est elle qualifie de periodique?
-indique quels sont les element qui ont des proprietes voisine de celles de l oxygene et idem pour celle du carbone
-comment mendelïev aurait il du placer les elements suivant berylium magnesium calcium et stontium
- peut on envisage la synthese de nopuveaux elements par sdes reaction chimiques ?


MERCI D'AVANCE POUVEZ VOUS ME REPONDRE A BONOIM@YAHOO.FR


2656 -
modify delete 3431 - Reply from hamid , 34 yrs (algerie) - 2005-09-27

je peux rependre a toutestes questions et je peux t'aider dans tes recherche juste contacte moi

modify delete 2639 - from Clarise , 19 yrs (china) - 2005-04-14
Physics - "chemistry student"

Hello,I'm a university girl studying chemistry.If you are also studying chem and your native language is English,write to me:)
We can talk about chemistry as well as other things.
Of course if you have any problem with chem,you can ask me for help,though my chem isn't excellent enough.I will like to help you.
clarise@126.com


modify delete 2629 - from Ross (Australia) - 2005-04-12
Physics - "Addig/Subtracting Vecotrs & Relative Velocity"

Hey everyone, I am so confused with this it's not funny. This is my first time accessing a site like thisss, but I am desperate seeing as though it's school holidays atm where I am and I can't talk to my physics teacher for help. Anyhow, I hope one of you guys can help me out. :smile:

Ok I have 2 problems.

One is with adding and subtracting vectors. Ok when I have to vectors to add or subtract say

12ms-1 west + 14 ms-1 north

< IMG SRC="http://members.optushome.com.au/rossverg2002/1.JPG">

Before I do anything I draw my diagram as descbried above. The problem is that in oder to add them I need to have the vectors touching arrow-to-tail. The problem is which one of the arrows do I reverse and how do I express the new velocity (with a - sign?). Does it even matter which arrow (velocity and direction) I reverse?

Ok second problem :frown: ... I am confused with relative velocity. In my textbook there is a formula:

Velocity of a relative to b = velocity of a - velocity of b.

How do I know when to use the formula or vector diagrams? Is it only meant for velocities in a straight line? What do I do when there are 2 dimensional velcoities (like a velocity north relative to a velocity east); do I use vectr diagrams instead of teh fomrula??


Thankyou to all who help.
Your help is greatly appreciated. o:)


modify delete 2622 - from sean (belgium) - 2005-04-11
Physics - "deriving these equations"

how do i derive G(v) = we (v + 1/2) - wexe(v +1/2)2

and v (wavenumber) = Te' - Te'' + G(v') - G(v'')

to give w'e - 2wex'e(v'+1)?

how is it possible to derive an equations that have parameters in both excited state AND ground state?


modify delete 2586 - from kranti , 19 yrs (india) - 2005-04-02
Physics - "magnetic field."

can we make earths magnetic field stronger by artificial means?
can our communication signals sent to satellites by magnetic fields?


modify delete 2564 - from jade , 16 yrs (france) - 2005-03-27
Physics - "Exercice de physique chimie"

serai t-il possible de me dire comment on calcule le poids et la force gravitationnelle. Ou alors pourriez-vous me donner les formules avec des explications si possible.
Merci d'avances


2564 -
modify delete 3220 - Reply from marco , 16 yrs (madagascar) - 2005-08-30

nen calcle la force gravitationnele par la formule de newtone
fA/b=GDISTANCE AU CARée sur masse de afois masse de b

modify delete 2556 - from becca (ireland) - 2005-03-25
Physics - "physics"

Describe how u wud plot d magnetic field of a bar magnet.
explain how u wud magnetise a piece of iron,


2556 -
modify delete 2566 - Reply from Eric , 19 yrs (cameroon) - 2005-03-27

There are many means to magnetise an iron .U can :
-set a wire round it and let a current circulate within it .or
-scrub a piece aof magnet on it .Some particle of the magnet will stay on it ,or
-thta's not the best way but u can also hot it.
Concerning the plotting of the mag field ,it's as all the mag fields .they are lines going from one extremity to the other ,directed north-south

modify delete 2543 - from Dilan , 17 yrs (Australia) - 2005-03-21
Physics - "Motion down an incline"

Hi Guys,

For a physics practical investigation I will be investigating the change in height (angle) of an inclined plane and the time taken to travel down the incline.
Now for my hypothesis and results I am a bit stuck. For the hypothesis I will be using the following formula.

x=ut+.5at^2

Where:
x=distance down the incline plane (meters)
u=Initial velocity (m/s)
t=time taken to travel down (seconds)
a=acceleration down the incline plane (m/s/s meters per second per second)

Now for our situation
x=1.58
u=0
t=?
a=gSinΦ (Φ varies with different heights)

So now I can build a relationship between the height (or the angle of inclination) and the time taken to travel.

Now I substitute values into the formula to get:

x=.5at^2 (because u=0)

Then transpose:
2x=at^2
t^2=(2x)/(gsinΦ)

Now we know that the values of 2x and g (g = 9.8 = acceleration due to gravity) remain constant (well we assume they do). How do I create a direct proportionality between t and Φ ?? So I get a straight line graph when I plot the two?


2543 -
modify delete 2573 - Reply from helper (USA) - 2005-03-29

Unfortunately, there is no real way to make a straight line or direct relationship between the two. This is arising from the fact time will be proportional to the square root of a transidental function (cosecant = csc = 1/sin). Even if you try more advanced mathematic (say, complex number theory), you can not make them directly proportional based upon your equation. Sorry.

My best recommendation is to t vs sqrt(csc(phi)) for various phi's and see what kind of plot you get. I tried it using Excel and found a curve that looks like y=a/x, where a is a constant, until phi gets 'too large'. This is to be expected as, for small angles only, sin(phi)~=phi.

Wish I had better news!

modify delete 2514 - from Pam (US) - 2005-03-11
Physics - "Help!"

What would be the final temperature of 100g of 20 degree C water when 100g of 40 degree iron nails are submerged in it? (The specific heat of iron is 0.12 cal/g C. Here you should equate the heat gained by the water to the heat lost by the nails). How do I figure this?


2514 -
modify delete 2551 - Reply from gyani , 25 yrs (India) - 2005-03-23

i can answer your question, but i like to use diagrams for my explanations to make it easier to understand. I'm wondering if there is a way to attach diagrams here.???

modify delete 2460 - from Joe , 12 yrs (US) - 2005-02-28
Physics - "Lawrencium Properties"

I need to know some properties, or facts about the element Lawrencium. I already know that it is manmade, discovered in 1961 by Alberto Ghirro, and named after Ernst Lawrence, that its color, density, freezing point, melting point, weight, and uses are unknown. and I also know that it has 103 p+ 103 e- and 159 n.


modify delete 2459 - from Joe , 12 yrs (US) - 2005-02-28
Physics - "Lawrencium Properties"

I need to know some properties, or facts about the element Lawrencium. I already know that it is manmade, discovered in 1961 by Alberto Ghirro, and named after Ernst Lawrence, that its color, density, freezing point, melting point, weight, and uses are unknown. and I also know that it has 103 p+ 103 e- and 159 n.


modify delete 2448 - from Erin , 18 yrs (USA) - 2005-02-25
Physics - "Impossible Physics problem i need help on"

ok the formula for this problem is Fg= G*(m*m/d^2) Where G= 6.67x10^-11

the problem is:

a spacecraft is on a journey to the moon. The distance between the center of the earth and the center of the moon is 3.85x10^8m. At what point, as measured from the center of the earth, does the gravitational force exerted on the craft by the earth balance the gravitational force exerted by the moon? this point lies on the line between the centers of the earth and the moon


2448 -
modify delete 2449 - Reply from mh (germany) - 2005-02-25

Just use the formula: the gravitational force of the earth must be equal to the grav. force of the moon in this point: F_earth=F_moon --> G*m_earth*m_spacecraft/d^2=G*m_moon*m_spacecraft/(D-d)^2 and D is the whole distance between the surfaces! earth-moon (3.85*10^8m-radius_moon-radius_earth) the equation can be simplified because you can "cancel" both sites with G and m_spacecraft (so the result must be independent of the mass of the spacecraft!!) so m_earth/d^2=m_moon/(D-d)^2 then you transform this to a quadratic equation - I show every step... 1) ((D-d)^2/d^2)=(m_moon/m_earth) 2) (D^2-2*d*D+d^2)/d^2=(m_moon/m_earth) 3) (D/d)^2-2*(D/d)+(1-(m_moon/m_earth))=0 this is the quadratic equation which can be solved by the pq-formula... - there are two mathematical results but one of them doesn't describe the situation on the direct connection between earth and moon: 1) d=D/(1+sqrt(m_moon/m_earth)) 2) d=D/(1-sqrt(m_moon/m_earth)) so only the first one is a valid result. Now you can insert the values for m_moon=0,07349*10^24[kg],m_earth=5,9736*10^24[kg],r_moon=1,738*10^6[m],r_earth=6,378*10^6[m] to calculate: D=3.85*10^8[m]-1,738*10^6[m]-6,378*10^6[m]=3.768*10^8 with the value of D you can calculate d=3.39*10^8[m] but in your question the value of the point is measured from the center of earth so you add to this result the radius of the earth: d_result=3.39*10^8[m]+6,378*10^6[m]=3.455*10^8[m].

modify delete 2435 - from Journey (USA) - 2005-02-21
Physics - "Kinetic energy"

How far would a 4 kg book fall to pick up a kinetic energy of 80 Joules?


2435 -
modify delete 2451 - Reply from mh (germany) - 2005-02-26

to my mind guidoo makes one little mistake - the mass of the book is 4kg but he inserts 80 kg in the formula to calculate the energy - so E = 1/2 * m * v^2 = 1/2 * 4[kg] * v^2 = 80 J and v=sqrt(40) then you come to the right result...

2435 -
modify delete 2436 - Reply from guidoo (netherlands) - 2005-02-22

use the formula:

E(kinetic energy) = 1/2 * m(mass)* v(velocity^2

fill in your given conditions:

E = 1/2 * m * v^2 = 1/2 * 80[kg] * v^2 = 80 joules

which gives:

v = sqrt(2) = -+ 1.41

the speed of the book corresponding to 80 joules is about 1.41 [m/s]



now this doesnt give us a distant yet. what we _can_ do is find out how long it takes for a book to accelerate to 1.41 [m/s] when you drop it. use the formula:

(end velocity) - (starting velocity) = a(acceleration) * t(time)

you can look up the earths g-force (gravitation), which is 9.8 [m/s^2]
fill in:

1.41-0 = 9.81 * t

which gives t =1.41/9.8 = 0.14 seconds

this means that our book has reached the speed of 1.41 m/s and thereby the kinetic energy of 80 joules, in 0.14 seconds.


finaly, we use the following formula to relate distance, time and acceleration (gravity in our case):

s(distance)=a(acceleration) * t^2 (time squared)

fill in:

s = 9.8 * 0.14^2 = 0.19 meters

our answer is 0.19 meters!

modify delete 2418 - from kitty (canada) - 2005-02-18
Physics - "entropy"


Hi,
i'm having trouble with a thermal physics problem relating to the partition function and i was wondering if anyone could help me out. the problem is as follows:

(a) Consider a molecule which has energy levels En=c|n| , where n is a vector with integer components. Compute the partition function for a
one-dimensional ideal gas of such molecules. Show that the average energy (assuming large temperature compared to the difference between adjacent energy levels) gives U=N(kB)T.

(b) Take c=0.5eV. For a lab-sized box of gas of our molecules, estimate the temperature (in Kelvins) at which the assumption of part (a) breaks down.

(c) Find the entropy. Figure out if this expression breaks down at low temperature. Are the two breakdown temperatures of (b) and (c) similar? Why or why not?

----------------------------------
For part b), I guess the assumption is that the temperature, t (in fundamental units), is much greater than the change in energy levels, (E(n+1) - E(n)). As i understand, this assumption was used so that the summation in the partition function can be approximated by an integral.

Let T be temperature in kelvins and kB be the boltzman constant. I got,

t >> E(n+1) - E(n)
t >> c|n+1| - c|n|
t >> c
t >> 0.5eV
(kB)T >> 0.5eV
T >> [0.5eV * 1.602 *10^-19 J/eV] / kB
T >> 5802.26 Kelvins

Is my approach for doing this problem correct?

----------------------------------------

I'm mainly stuck on part c). let s = entropy (in fundamental units), F = free energy, t = temperature (in fundamental units) and Z = partition function. let N = # of gas molecules in the box.

I found the entropy as follows:

from part a), i found Z to be 1/N! * (t/c)^N

s = - dF/dt
= - d/dt [-t ln(z)]
= d/dt [t ln(1/N! * (t/c)^N)]
= d/dt t[ln(t^N) - ln(c^N) - N ln(N) + N] --> by stirling's approx.
.
.
.
= N [ln(t/cN) +2]

I don't know where to go from here. how does this expression break down at low temperature? It seems like the only temperature that doesn't work is 0 and also negative temperatures, since you can't take the log of 0 or negatives. but t can't be negative anyway because we're using fundamental units (eg. NOT the celsius scale). how would i find the breakdown temperature in this case? is it supposed to be different than the answer in part b)? I understand why we must assume a large temperature in part a) (t/change in energy >> 1) - it's so we could approximate the summation by an integral when we're finding the partition function. But I don't understand why we need to assume a large temperature when finding the entropy. I would appreciate any help. thanks.


modify delete 2401 - from papz , 17 yrs (engand) - 2005-02-15
Physics - "Help on a chem question"

--------------------------------------------------------------------------------

HA(s)+NaOH(l)= NaA(aq)+H20(l)


7.1g of acid was added to 250cm^3 of water in a 250cm^3 flask of water, then 25cm^3 of it is placed in the buiret via a pipette. On average it takes 23.27cm^3 of the HA to change the colour of 25cm^3 of 0.15M NaOH to change colour.
What is the concentration of the HA, also the manufacuture says the purity if 99.1%, find your value of the purity and comment on differece (if any)




i posted one like this b4 but had got some of the data mixed up, anyone help me?


modify delete 2400 - from Charles (France) - 2005-02-15
Physics - "Dm de physique de premiere S tres tres dure"

j'ai recopié les questions si vous voulez:POUR LES SCHEMAS, VOIR LE LIEN http://exos.site.voila.fr/page1.html

Etude d'un ballon sonde
Ce problème porte sur l'étude d'un ballon-sonde, constitué d'une enveloppe sphérique E, gonflée à l'hélium, rigide, de rayon constant R=12.0m et de masse négligeable devant celle de la nacelle N, de masse m=5.00t, de volume négligeable devant celui de l'enveloppe.

1. Equilibre de l'enveloppe au sol
Initialement, l'enveloppe, sans nacelle, est retenue au sol par deux câbles sans masse, attachés au sol en deux points A et B. Les câbles font avec le sol des angles alfa1=40.0° et alfa2=50.0°.

1.1 Faire le bilan des forces s'exerçant sur l'enveloppe.

1.2 Déterminer les intensité de toutes les forces appliquées a l'enveloppe.

Ascension du ballon

On fixe la nacelle N sous l'enveloppe E, et on largue les amarres AA' et BB'.

2.1 Faire le bilan des forces appliquées au ballon (ensemble nacelle + enveloppe)

2.2 Calculer l'intensité F de la résultante vecteurF des forces au niveau du sol. Le ballon reste-t-il immobile ?

Au fur et à mesure de l'ascension du ballon, l'air se raréfie et sa masse volumique p diminue. Arrivé à une certaine altitude Zm, le ballon est en équilibre.

2.3 Calculer l'intensité Pie de la poussée d'Archimède exercée sur l'enveloppe à l'altitude Zm.

2.4 En déduire la masse volumique p de l'air à cette altitude.

2.5 A l'aide du schéma2, donnant les variations de la masse volumique p de l'air en fonction de l'altitude, déterminer la valeur de l'altitude maximale Zm atteinte par le ballon.

Descente de la nacelle

Une fois l'altitude maximale atteinte, la nacelle est larguée, et descend seule avec un parachute. Celui-ci exerce une force de frottement vecteurf dont l'intensité j=k*v(au carré) est proportionnelle au carré de la vitesse v de chute du ballon. Très rapidement, le ballon tombe à vitesse constante Vo.

3.1 Faire le billan des forces s'exerçant sur le ballon lors de la phase de chute à vitesse constante. La masse du parachute est négligée devant celle de la nacelle.

3.2 Déterminer la valeur de la vitesse Vo, sachant que le coefficient de frottement k vaut k=5.5*10(puissance3)kg*m(puissance-3). Comment s'appelle cette vitesse ?

3.3 Calculer l'énergie cinétique Ec de la nacelle au cours de la chute.

3.4 La nacelle tombe d'une altitude Zo=4000m. Quelle est son énergie potentielle de pesanteur Epp en début de chute ? L'origine des énergies potentielles est prise au sol.

3.5 Calculer les variations deltaEc d'énergie cinétique, deltaEpp d'énergie potentielle de pesanteur et deltaEm d'énergie mécanique entre l'altitude Zo et le sol.

3.6 Quelle est la quantité Q de chaleur fournie à l'atmosphère au cours de la chute ?

DONNEES:
Pesanteur terrestre: g=9.81N/kg(puissance-1)
Volume d'une sphère: V=4/3*pie*r(au cube)
Masse volumique de l'air au niveau du sol: o=1.19kg*m(puissance-3)





2400 -
modify delete 2563 - Reply from otman (japan) - 2005-03-27

what a fool its so easy

















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