space probe question-Math?
912921
altitude = height = 552km\nAngle of dep = 39.7\n\nTangent of Angle = Alt/ Radius\n\nRadius = Altitude/ Tan(39.7) = 552/tan(39.7) I think... p.s not really sure
248915
The limit as sky goes to zero from the positive direction would be ground.\nLim    = ground\nsky-->0+\n\nThe limit as sky goes to zero from the negative direction is space.\nLim    = space\nsky-->0-\n\nSince the limit from the + site does not equal the limit from the - side, the limit does not exist.
732740
im curious how this helpful to both of us??\n\nlynns distance (neglect directions )is 20 km....but her displacement (a vector) is 2 km east....because going north 5 and them south 3 and south 2 later cancels out n/s distance and going 6 east but 4 west leaves with 2 east..draw it out\n\n\nthe UFO moved 10 km North, it moved 20 east but them 30 west, subtract to leave 10 West...so the TOTAL displacement is 20 km\n\nBUT use pythagorean for 2\n\n10^2+10^2=200, sqrt 200, leaves 14.14 km NW
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
1135263
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.
344788
hey man, dont worry ill help. first of all our given:\nh=280m\nsg=1.03\ngc=9.7m/s^2\n\nreq'd Pressure(absolute)\n----------------------------------------------\nsol'n:\nfor DENSITY of subs:\nformula\nSpecific Gravity=density of subs/density of water\ndensity of subs=1.03 x 1000(density for water)\ndensity of subs=1030 kg/m^3\n\nFOR PRESSURE GAGE:\nP(gage)=Vh\nP(gage)=density x gravitational constant x height\nP(gage)=1030 x 9.7 x 280\nP(gage)=2,797,480 Pa\n\nFOR PRESSURE ABSOLUTE:\nP(abs)=P(gage) - P(atm)\n\nwhy minus? bec. the submarine is pressurized, meaning its vacuumed\n\nP(abs)=2,797,480 Pa - 1atm(101.325x10^3 Pa)\nP(abs)=2,696,155 Pa\n\nconvert Pa to Kpa\n=2,696,480 /1000\n=2,696.155 Kpa\n\nconvert to bar\n=2696.155kPa x (1.01325 bar / 101.325 kPa)\n=26.96155 bar\n\ndont mind the others, they dont even know how to answer this simple question. They just want to earn points!
697340
first: The mass of the satellite doesnt matter.  it NEVER does in orbit problems.  No really knows why beyond that the m's cancel.\n\nT^2=4 pi^2 r^3\n       -----------\n            GM\n\nr= radius of the orbit(radius of the earth)\npi= 3.14\nG= gravitational constant= 6.67E-11 ..i think...\nM= mass of earth\n\nthe answer is .0581 days or 83.7 min.
767928
1. its final velocity depends on the direction it was heading to start off with. If it was flying in the x direction (no components in y or z) then after 4.3 seconds the x component has increased by 8.92*4.3  and the y component has increased (from 0 this time) by 1.21*4.3.\nso now you can work out the new direction (in the x direction but slightly downwards by whatever angle has a tangent of 1.21*4.3/(152+8.922*4.3) \nand using pythagoras to determine the speed.\n\n2. You can ignore the gravity as it is << the precision with which the acceleration due to the electric field is stated (I'm assuming the 106 is 10^6 = 1000000) \nso the final velocity is 3.80*2.45*1000000 m/s upwards.
1175686
This can be solved in several different ways.  One of course is by calculus, by developing a formula for the surface area of a "slice" of the remainder of the sphere and integrating it about a rotated axis, or something similar.  Another is by digging for the rather obscure formula for the "volume of the endcaps of a sphere" and subtracting two of those and the volume of a cylinder from the volume of the original sphere (and discovering that many of the terms will cancel out nicely).\n\nThe easiest method, however, is via the use of simple logic.  Unfortunately, this method also requires one more "given" in order for it to be valid, though this given is not an unreasonable one:  There must be a single, real answer (or more specifically, a unique answer on the real number line).  Such an assumption would be folly in a practical (real-world) context, since without some form of proof, one should never immediately assume that this is the case.  Perhaps there are multiple answers, or there is no answer at all?  But as luck would have it, there actually is one answer, and it may be derived logically due to its unique existence.  (*thumbs nose at any philosophers who may be reading*)\n\nSolution:\n\nLet R be the radius of the cylinder drilled into the sphere.  If and only if there is a single, real answer, then that answer would still be the same no matter the value of R... even if R=0.  Thus, the answer is the volume of a sphere of radius 3 (diameter 6), or 36*pi cubic units.\n\n(Yes, the volume is always the same, whether the original sphere were the size of a basketball, or the size of a planet.)\n\nIt's a surprising answer, along the lines of the more familiar puzzle: "If the Earth were a perfect sphere and a band were placed snugly about its equator, and then one more meter were added to the length of the band and it were stretched so that all points along it were equidistant from the surface of the Earth, how far above the Earth would the band rise?"  Many are surprised to find that the size of the Earth plays absolutely no part in the equation, and that the "rise" is exactly the same if the sphere were basketball-sized or, umm... Earth-sized (with a circumference greater than 40,000,000 meters).  It rises exactly the radius of a circle with a 1-meter circumference: 1 / (2*pi) meters.
768643
Here you go: http://www.gsfc.nasa.gov/scienceques2004/20041105.htm
509966
Ummm that's not a real question.\n\nBut, if that happened, you'd end up with someone standing on the earth who observed that small particle moving in a direction opposite the earth's rotation at a speed that could be calculated by:\n\n(radius of the earth)*6.28*(1/30-1/24)
700031
You are looking for an ellipse.  The inside of the ellipse would be the area that the scout could survey.  20 would be the distance between the foci and 40 would be the vertices.  Good Luck
307097
Ksp = [Ag+]^2[CrO4 2-]\nsince [Ag+] = 2*[CrO4 2-]\nKsp = 2*[Ag+]^3\nsolve for [Ag+] then plug it in this equation\n[Ag+] = mass /0.010L H2O
547016
This is a 3D problem that can be treated as a 2D problem due to azymuthal symmetry.\nFirst of all recall the general equation for a parabola:\ny-y0 = 4*p (x-x0)^2\nWhere p is the focal lenght.\nNow, if we set the vertex of the parabola (in this case of the parabolic dish) at the origin, x0=0, y0=0.\ny = 4*p x^2\nIn the previous equation susbtitute y=3, x=5.\nThe resulting 'p' will be the focal length measured in feet.
754826
The distance required for the asteroid to travel around the sun = \n\n2PI *radius.....Radius R = 632 million kilometers\nThe orbiting circle for the asteroid is 3968.96 million kilometers\n\nDividing by the time it takes the asteroid to orbit a whole orbit around the sun, we should get the velocity\n\n3968.96 / 8.6485 million kilometers per year.\n\nOne million kilometers multiplied by 10^9  converts to meters\nOne year is = 365*24*3600 = seconds \n\nBy converting the Kilometers to meters and the years to seconds you should get:\n\n1257312021 / 11364129 = \n\n14552 meters per seconds approximately\n\nYou should double check the divisions and multiplication.
752738
The 120N force on the astronaut accelerates him/her at a rate of 120/79 m/s^2. In 0.42 s, this would accelerate the astronaut to a velocity of 0.42 (120/79) m/s.\n\nSimilarly, the satellite accelerates in the opposite direction to a velocity of 0.42 (120/660) m/s\n\nThe relative velocity of the astronaut and the satellite is \n0.42 (120) (1/79 + 1/660) m/s\n\nMultiply this by 84 seconds (which is equivalent to 1.4 min), and you'll have the requested distance. \nNote: This is the distance the astronaut will be from the satellite IN ADDITION TO the length of the astronaut's arms. We started counting the time when the astronaut had straightened his/her arms.
922469
I found this very easily on the Internet.  You could have done the same thing.\n\nhttp://liftoff.msfc.nasa.gov/academy/universe/RADEC.HTML