How is Ksp affected by this?
371354
The Ksp for a salt is a constant and should not change in response to solution volume.  However, we must be careful not to forget the definition of Ksp.  The Ksp of an alkaline earth hydroxide that dissociates according to the following reaction is:\n\nX(OH)2 <----> X(+2)  + 2OH(-1)\n\nKsp = [X][OH]^2\n\nIn other words we simply multiply the concentration of the metal ion by the concentration of the hydroxide ion squared, but ONLY IF THE DISSOCIATED IONS ARE IN EQUILIBRIUM WITH SOME UNDISSOCIATED SOLID!!  So there must be some undissolved solid that remains.  If we add so much solvent that the solubility limit of our salt is not reached, and it dissolves entirely, then we can NOT make any measurement of the Ksp.\n\nIn short the answer is:  This will have no affect on the calculated Ksp as long as the additional volume of solvent added did not cause all of the salt to dissolve.
538789
The fact that there's pretty much nothing random in orbital mechanics is very helpful along with the fact that we know pretty much everything about what affect all the objects in the solar system have on the spacecraft, that way they can just run the simulations using computers to high precision and take account of pretty much everything.\n\nIt is of course common to use mid-course correction burns to correct for any errors when the craft was first sent off into space.
275035
There is more friction on re-entry. Friction is geometrically proportional to speed. The shuttle takes a long time to accelerate into orbit, but it comes back full speed. Also the flight angle is diferent on re-entry, exposing more surface.
348312
as someone pointed out. it could be because of a very large angular rotation decreasing the air resistance.
981333
http://www.gsfc.nasa.gov/scienceques2002/20030516.htm
251863
At 20 degrees C, it's 0.6g/100mL. You can convert that to a Ksp.
478657
nothing.\n\nthere is no documented effect of this kind.
1077675
affects the structure???  how would it affect the structure???  the performance, yes... have you even done anything besides ask dumb questions on a computer?
315574
I think this is how you do it:\nPb + F --> PbF\n\nSo Ksp = [Pb][F]\nBecause they react in a 1 to 1 ratio, their concentrations will be the same.\nSo Ksp = X^2 or [Pb]=[F]=(3.2E-8)^2\n-Not 100% sure but I think thats it
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
273733
Yes\n\nRemember:\n\nKsp = [Cation][Anion] (concentrations) / [Solid] (always 1)\n\nAs the concentrates of cation and anion decrease, their product decreases (making KSP smaller)
999938
One of the biggest risks in space is exposure to radiation. Highly-energetic particles - called cosmic rays - are constantly zooming through space. These particles, when they interact with the cells in our body, can cause these cells to mutate and form cancers and tumors. The same thing can be said for the other forms of radiation, most of which come from the sun but some of which comes to us through interplanetary space. These are the ultraviolet and infrared rays given off by the sun, as well as other atomic and nuclear particles which have the same affect as the cosmic rays mentioned before. These radiation also can adversely affect any electronic equipment - satellites, space shuttles, etc. We are protected from a majority of these rays and particles by the Earth's atmosphere; there is no such protection in space.\n\nAnother consideration is the bits of dust that are drifting about out there. These grains (micrometeorites) are extremely tiny but have a velocity which can - sometimes - penetrate a spacesuit or ship. Getting hit with one probably won't be terrible but, a few hundred can severely damage a spacesuit, shuttle or whatever equipment might be lying in their path.
989461
parachute is affected by two forces: weight and air resistance\nm.g weight is fix\nK.v² air resistance is proportional to:\n-square of speed\n-viscosity of air (eps)\n-projection of area of contact (A)\n-form coefficient (alfa)\nalfa is related to the dynamic for of the contact surface\nalfa is minimum [ex: 0.06] for best air design [like bullet]\nalfa is 1 for plan surface\nalfa is 6 for hemisferic surface\ndynamic relation:\nm.g-K.v²=mv';    v' is acceleration, which is derivative of speed\nv starts with 0 and m.g-K.v² starts with mg - 0=mg >0\nmv' starts =mg >0\nspeed increases as very quickly\nv increases so m.g-K.v² decreases\nso mv' decreases and the acceleration is decreasing, speed v is still increasing but slowly\nv continues to increase for several seconds until m.g-Kv² becomes 0\nwhen that comes, the acceleration becomes 0 and speed is fix = maximum speed.\n\nthat maximum speed is Vmax:   m.g-K[Vmax]²=0\nVmax=sqrt[m.g/K]\nthat is the speed of falling of parachute\nVmax=sqrt[m.g/(eps.A.alfa)]\nVmax is invers proportional to sqrt of projection of surface contact and sqrt of form coefficient.\nif the parachute is with same form, same surface contact its speed will not be affected by no. of strings neither by diameter. They almost have no contact with air
529130
What is it about launches that you consider anomalies...\n\nPlease e-Mail me, if you wish, at ChrisMGarrett@juno.com
743571
Human space junk can hit satellites such as the Hubble telescope. The same is true for small rocky debris floating thru space. \n\nThe old Skylab came crashing to earth in the vicinity of Kalgoorlie, W. Australia, causing all sorts of Chicken Little jokes, "the skylab is falling, the skylab is falling". No one was hurt, by the way. Other stuff has come to earth too.\n\nSpace junk does not really affect the sun or magnetic fields.
91362
It's tranferred to heat/friction from the impact
955261
i dunno, cuz it's kewl, mebbe?\nBTW, imho, this NASA thing is a boondoggle