who set it.\nwhat time\n\nif u can give the altitude in meter. I would really appreciate it
125181
Highest I have done is 16,000 ft but ... \n\n"Kittinger performed three extreme altitude jumps during August 1960 as part of the USAF research project ‘Excelsior’ – a precursor to the US space race and intended to test human survivability on the edge of space. \n\nExcelsior III was the climax of the operation. \nLeaping from the open gondola of a 200ft diameter helium balloon with 99% of the earth's atmosphere below him, Kittinger spent four minutes thirty-six seconds in freefall. As he passed through 90,000ft his speed reached an incredible 614mph – almost the speed of sound in thin freezing air of the upper atmosphere. At 18,000ft his multi-stage parachute, operating on timers and barometric triggers, finally deployed his ‘main’.\n\nThis historic jump set an altitude record that still stands today."
416506
When the balloon is 30 meters above the ground, you can use a right triangle to find the distance from the balloon and the observer.  Since the height is 30 meters and the distance from the observer to the balloon on ground is also 30 meters, you have a right triangle with both legs = 30 meters.\n\nYou can now deduce an equation to relate the sides of the triangle with the angle of elevation.  Using simple trigonometry, you have:\n\ntan(x) = (height)/(ground distance)\n\nTaking the derivative of both sides with respect to 't', you get:\n\nsec^2(x)(dx/dt) = (1/ground distance)*(dh/dt), and we know that sec(x) = 1/cos(x).  Therefore, this can be simplified to:\n\ndx/dt = [cos^2(x)]/(ground distance)*(dh/dt).  Since we do not know the angle 'x', we can re-write the cosine relationship using trig.  We know that:\n\ncos(x) = (ground distance)/(hypotenuse).  The hypotenuse is found using the 45:45:90 special right triangle.  In this relationship, the sides of the triangle have the relationship x:x:x*sqrt(2).  Therefore, we can conclude that the hypotenuse is 30*sqrt(2) = 42.43 m\n\nNow you can simplify:\n\ndx/dt = (ground distance/hypotenuse)^2/(ground distance)*(dh/dt)\n\n= (ground distance)/(hypotenuse)^2*(dh/dt).  Plug in our known values and solve for dx/dt\n\ndx/dt = (30)*(3 m/s)/[(30*sqrt(2))^2]\n\ndx/dt = 0.05 degree/second\n\n------------\n\nHope this helps
500470
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
342174
You don't own the meter ... Con Ed does.\n\nCall them and they will do it.
348886
If you drop a vertical line h representing the balloon's altitude, you have two right triangles.  Let the base of the 83 degree triangle be x.  Now here's where you run into a question and if the book shows a diagram, it might answer that question.  Are the stations looking toward a balloon in between them, or are they both looking in the same direction at a balloon very much farther away, and very much higher?  If it's between them, the base of the 57 degree triangle will be 15 km-x, but if it's on the same side of them both, that base will be 15 km+x.  Do you know the trick to remembering how the sides relate to the trig functions?  Ooh--aah--oh ah--ah oh--ha--ho.  You have to sound this out and it makes a distinctive but stupid sounding little chant.  For the sides opposite an angle, adjacent to the angle and the hypotenuse in a right triangle, sin=o/h, cos=a/h, tan=o/a, cot=a/o, sec=h/a, csc=h/o.  In this case we want the ratio of the height to the base, the opposite side being the height and the adjacent side being the base, so we want the tangents of the angles.  The altitude, opposite side to the angles, will be equal for both triangles.  Call it y.  x will be the base for the 83 degree triangle in both cases, but for the 57 degree triangle the base will be (15 km-x) if the balloon is between the stations or (15 km+x) if it's to the same side of both stations.  Tan 83=h/x and tan 57=h/(15-x) (between) or tan 57=h/(15+x) (same side).  Setting the heights of the triangles equal since the balloon can only be at one height: x tan 83=(15-x) tan 57 if between, x tan 83=(15+x) tan 57 if to the same side.  Solve for x.  Then plug x back into the earlier equations to get y, the altitude.  Work out the answer on your own, but I'll give you this: if the balloon is between the stations, x is between 2 and 3 and y is between 15 and 20.  If the balloon is to one side of both the stations, x is between 3 and 4 and y is between 25 and 30.
89539
16UT? check your universal time.\n\n----Addition----\n\nI agree with the answer below.\nFor 11º 12' N :  60º 44' W, zone time is -4, so you are trying to calculate azimuth at 12:00pm (16UT), this gives an azimuth answer of 166.1 and altitude of 81.7\n\n255.9º at 12:00pm is too much in east direction.
613291
hi..yea i got a calendar/clock on mine..its in military time but its pretty kewl cuz it also tells the date and all that..im not sure but i think that the place that i got this calendar/clock also has just plain clocks, let me give u the link so u can look around the site\nwww.123mycodes.com
63077
2 hours. AZ time always confuses me.
408606
Blame it on the Brits:\n\n"In 1792, the British East India Company established the Madras Observatory in Chennai (then Madras), largely due to the efforts of the British sailor-astronomer Michael Topping. In 1802, John Goldingham, nominated as the first official astronomer of the Company in India, established the longitude of Madras (13°5′24″N, 80°18′30″E) as 5 hours and 30 minutes ahead of Greenwich Mean Time as the local standard time. This marked the first-ever instance of the use of the current time zone, and departure from the earlier standard of day-beginning at sunrise getting shifted to midnight."\n\nIt is considered Indian Standard Time and the full explanation can be found here:\n\nhttp://en.wikipedia.org/wiki/Indian_Standard_Time\n\nThanks for a good search!!
575631
Right ascension and declination are just like longitude and latitude, but in the sky instead of on Earth. The stars are fixed on this grid, so if you know the right ascension of any star then you can measure from that star to any other point. If you want to know the right ascension of the meridian at your location, or the hour angle of the vernal equinox, which is 0hr, you need to know the local sidereal time. The US Naval observatory has an online calculator. Of course local sidereal time is changing continuously as the Earth rotates, so at the moment you determine the local sidereal time you need to start a clock and later add whatever that clock says to the value you originally got to have the current local sidereal time. Also, the clock should really be a sidereal clock and not a solar clock, because a solar day is 24 hours but a sidereal day is only 23h 56m 4s, although a regular solar clock is probably good enough for a few hours. The difference is due to the Earth's orbit. Earth rotates once in 23h 56m 4s but in that time it has moved around the Sun enough so that it needs to rotate another 3m 56s to bring the Sun back to the same altitude in the sky. Another way to look at it is that the sun moved 3m 56s in right ascension during the day.
477803
a)\nRate = Distance/time = (24000 ft)/(20 min) = 1200 ft/min\n\nb)\nTo write A(t), you know that the max. value is 24000 and the plane descends at a constant, linear rate of 1200 ft/min.  Thus, you can write a linear equation of this model:\n\nA(t) = 24000 - 1200t, where t = time (in minutes)\n\nPlugging in 0 (at initial time), the height is 24000 feet, which matches the fact in the problem.\n\nc)\nThough linear functions have infinite ranges and domains, this problem is limited to the realism of the problem.  The range is only limited from the maximium value (24000 ft) to the ground (0 ft).  You can't have a negative value for altitude because that would not make sense in the problem.  The domain is limited to the time in air (20 min).  You can't have a negative time and time after 20 min is irrelevant because the plane has already landed.  Summarizing:\n\nDomain: 0 <= t <= 20\nRange: 0 <= A(t) <= 24000\n\n--------\n\nHope this helps
55265
I send mine and the time stamp is PST (Pacific Standard Time). My friend, 50 miles west of me and actually living ON the Pacifc coastline has an EST (Eastern Standard Time) stamp on his. We are smack in the middle of the Pacific Time zone. And the actual times of day don't make sense either. Go figure, huh?
383741
i'll set it up for you\n\n200=100e r3\n\nnow do it yourself
426785
My techer said that scientists had 2 clocks that were accurate to like, nano-second, or more, and put one on a plane and flew it aorund the world and when it landed it was different than the other which didnt move
158400
Start by getting the 12:00pm GMT out of the question... if your longitude is 6h17m32.143sW of the Prime Meridian, then your time zone is an even -7 hours offset from GMT.\n\nThe more precise question would be, "At my location, solar time at solar noon GMT is 6:17:32.143am. My latitude is 44°44'45.694"N. Where am I?"\n\nTo convert the time reference to longitude, consider that the International Date Line (180°E & 180°W) is 12h0m0sW of the Prime Meridian. 6h0m0sW would be an even 90°W. \n\n6:17am yields a differential of 5h42m27.857s. Convert that to seconds (5*60*60 + 42*60 + 27.857), divide by 43,200 (12h*60m*60s) and multiply that by 180° to get your numeric longitude, then multiply the decimal portion out to determine your minutes and seconds of longitude.
43713
>>>> what is the altitude of a right triangle?\n\nI have several here, and I am at 12 metres above sea level.\nSeriously, you haven’t provided enough information. You need at minimum the length of 2 sides or a side and an angle as well as the orientation to be able to calculate height (altitude).\n\n>>>>>>if the equation is d=2r what is the d and the r what exectly \n>>>>>>>am I finding with this equation \n\nd = diameter. r= radius. You are finding either how far it is from one side of a circle to another (d), or how far it is form the edge of a circle to the centre (r).
497318
Temperature is temperature no matter what elevation or altitude you are at - no need to know it.\n\nIf you want to correct station pressure readings to sea level for a weather station, then it may be helpful. The most practical and a reasonably accurate method though is to simply set your barometer several times (making small adjustments) to the sea level pressure using data from a good surface chart during daytime on a nice day with a weak pressure gradient (when it'll be very similar over tens of miles). Here is a good site for observations the past hour:\n\nhttp://www-frd.fsl.noaa.gov/mab/metar\n\nThey say data are experimental and not always accurate blah blah blah - but I never saw an error in METAR listings for any of the stations nor their locations - though you have to be able to read METAR reports. E-mail me if you wish regarding this.\n\nWhatever the case, Topozone is good if you want your elevation for any particular reason:\n\nhttp://www.topozone.com