Why is the x-component of force negative in this problem?

[Alt+F4]

Maybe one more,
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp04


Question 10

help !...

 

[Alt+F4]

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su05

Qustion 22 , how come it is ln(2)*4*2*2 = 11.09 Sec i thought the resistor that i would plugged in would have been the EQuivalent of them all

 

[big man]

help !...question 10

Well this one has a straightforward association between the electric field between the plates, the potential difference and the distance. V=Ed

http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/parallelplates/parallel_plates.asp

If your book doesn't have any theory on this then just go to the above site or better yet you can search hyperphysics for more info.

Qustion 22 , how come it is ln(2)*4*2*2 = 11.09 Sec i thought the resistor that i would plugged in would have been the EQuivalent of them all

Well like you were thinking the capacitor is discharging through the resistors. However, it won't be discharging through R1. It will only discharge through R2 and R3. This is because the path along R3 is the path of least resistance. The open switch is like having an infinite impedance so current will not flow down a path of infinite impedance...instead it will flow through the path marked by R2 and R3. This means R2 and R3 are in series, which makes an equivalent resistance of 8 ohms. This is why you get 11 microseconds.

 

[Alt+F4]

Well this one has a straightforward association between the electric field between the plates, the potential difference and the distance. V=Ed

http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/parallelplates/parallel_plates.asp

If your book doesn't have any theory on this then just go to the above site or better yet you can search hyperphysics for more info.
Well like you were thinking the capacitor is discharging through the resistors. However, it won't be discharging through R1. It will only discharge through R2 and R3. This is because the path along R3 is the path of least resistance. The open switch is like having an infinite impedance so current will not flow down a path of infinite impedance...instead it will flow through the path marked by R2 and R3. This means R2 and R3 are in series, which makes an equivalent resistance of 8 ohms. This is why you get 11 microseconds.

we have no book, basically practice exams and lectures that are meanigless

 

[big man]

Like I said hyperphysics is a pretty good site at explaining most physics concepts so you might want to bookmark that site!

Also download emule plus http://sourceforge.net/project/showfiles.php?group_id=71866

Do a search for 'Physics for scientists and engineers'. It is a really good book that explains everything well. It's what I used in my first year.

The authors are Serway, Beichner and Jewett.