2 free charge + Q each placeda destance r, therd charge placed between find location,magnitude and sign?
968370
If the origin of the coordinate system is placed at the middle of the two particles positions are -r/2 and r/2. Two charges of the same sign will repell each other so the force has a positive sign and it's magnitude is given by (C Q^2) / (r^2) Where C is the Coulomb constant.
631876
Partial text of both is included in\n\nhttp://www.maths.tcd.ie/pub/HistMath/People/Newton/CommerciumAccount/CommAcc.pdf
810779
Let the two +Q charges be placed left to right with the distance r separating them.  Let a third charge of unknown charge q be placed a distance x to the right of the first charge.  Now, look at each of the +Q charges and sum the electric force acting on them.  For example, the left hand +Q charge experiences a repellent force from the +Q on the right, pushing it towards the left.  To balance this out, there must be an attractive force pulling it to the right, so the q charge is negative.  However, this is just an observation.  In order to maintain the integrity of the algebra, we should continue to assume that q is positive, but recall the negative result llater when we check our answer..  The proportionality factor for electric field strength can be ignored, because it will cancel out, so you can treat Q/r^2 as the value under consideration.  In this case, we need Q/r^2 + q/x^2 = 0.  However, because the +Q charges are equal in magnitude, I will make the assumption that the distance x = r/2, so Q/r^2 = -4q/r^2 ==> Q = -4q ==> q = -Q/4.  Perform the same force balance on the right hand +Q charge to verify.\n\nIf the initial charges had been unequal, x would have remained a variable, and the complementary distance (r - x) would have been used for the second force balance in order to form a system of equations.
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
900057
a Q1 since it is between 0 and 67\nb Q2 since it is between 67 and 88\nc Q3 since it is between 88 and 105
724411
Sure...\n\nfirst we want to combine like terms on each side of the equals sign.  Think of the equals like the balance point on a scale... you can do anything on one side as long as you match it (keep it balanced) on the other!\n\nso we want to combine the variables and the constants...\n\n8a - 3a + 4 = -a -4 + 2\n5a + 4 = -a - 2\n\nnow we want all the variables on one side, and constants on the other:\n\n5a -(-a) + 4 - 4 = -a-(-a) - 2 - 4\n5a + a = -2 - 4\n6a = -6\n\nnow we just divide by 6 to get a alone:\n\n6a/6 = -6/6\na = -1\n\nHope this helps! :)
112147
Ya if you ever need to find a percent of something multi. by 100 or just move the decemial 2 spaces to the right. That does the same thing as Multi. by 100.
923039
email me, calcu_lust@yahoo.com, i'll write out a nice solution.\n\nEdit: Go to my website, and aforementioned nice solution is typed out as a .pdf for you to download, with all the correct symbols and such:\nhttp://geocities.com/calcu_lust/yahooans.pdf
800717
(1) First one is copied wrongly by you                                            (2) Where is the equal to sign?                                                        (3)This equn. also cannot be solved                                              (4)-p/16+5=9  Multiplying all the terms by 16 ,we get -p+80=144   =>-p=144-80=66 Therefore p=66 YOU SHOULD BE MORE CAREFUL BEFORE SENDING YOUR PROBLEMS TO YAHOO ANSWERS.HOWEVER,YOU MAY SEND THE CORRECTED VERSION OF THE PROBLEMS TO ME IMMEDIATELY THROUGH YAHOO ANSWERS.
971561
free + 2 * twofield + 3 * threefield = 385\nfree = 3 * threefield + 11\nfree = twofield - 26\n\nso\n\nthreefield = (free - 11)/3\ntwofield = free + 26\n\nsubstitution method\n\nfree + 2 * (free + 26) + 3 * (free - 11) / 3 = 385\n\n4 * free + 41 = 385\n4 * free = 344\n\nfree = 86\n\nplug this back into the above to get other two variables, so\n\nfree = 86\ntwofield = 112\nthreefield = 25
743387
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
821637
1 break up the compuond into ions or ions and radicals in this case there are no radicals.\n2 place the charges of each ion to the upper right of the symbol and viola e.g. \nBa2+  Fl-
281772
You need 2 equals signs:\n\nif (toBeRemoved.left = null || toBeRemoved.right = null)\n\nshould be\n\nif (toBeRemoved.left == null || toBeRemoved.right == null)\n\nOne equals is for assignment, 2 for comparison.
1154468
make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
12600
In Custom format you can specify format as required by you.\nIt has three parts seperated by ;\n1st is for positive\n2nd is for negetive\n3rd is for Zero\nso you can specify as  follows to get the + sign displyed\n\n+#;-#;"Zero"
704060
let y=total charges\ny=$0.07x+$4.95
960662
ry +s = tx - m\nYou must leave y alone for indicate its value in terms of other variables.\n\nFirst of all you move the s to the other side or the = as shown: ry =tx- m - s (you change the sign of s)\n\nThen you move the r as follows: y=(tx-m-s)/r\n\nAnd thats it \n\nYou have a restriction because the r has to be different than 0 to avoid a division by zero.  So restriction is r different from zero.