a plane took of from 7 degree 53'6.1" and landed at 8 degree 6' 43.5'' it takes 12 hrs. find the local time
398878
same time as when it took off - you'll be in a different time zone????
925426
one half and seven eightths
768098
5 hours behind... \n\nThe co-ordinates belong to Toronto, Canada which is EST (Eastern Standard Time) & it is west of Greenwich
1174698
16 min approx by light\n250,000 hours by jet\n2 million hours approx by fast car\n80 million hours by foot\n\n150 Tm from earth (150 million km)
202077
blvckdrvgon is almost correct.\n\n1 degree is equal to 60 minutes (')\n1 minute is equal to 60 seconds ('')\n\nThe location of my house is 51º20'43''N, 00º07'34''W
1094919
Latitude does not affect time, it just affects daylight length. Longitude affects time\nNow ... the earth is a circle, (360 degrees) it turns once a day, 360 degrees in 24 hours, 15 degrees an hour. The 0 meridian is at Greenwich in London. so 10 degrees east is 40 minutes in front of Greenwich, and 70 degrees west is 280 minutes behind Greenwich. Using this we can say that 70 degrees west is 320 minutes behind 10 degrees east that is 5hrs and 20 minutes take that away from the flight time and you have 4 hours and 40 minutes, this means it will be 06.40 a.m. , local time, when you land
446591
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
63054
People inside the plane are still being pulled down by gravity with a force of m*g=9.8*m.  The airplane is also holding the people within it with a force of m*v^2/r=22.22*m.  So you've got a horizontal force and a vertical force.\n\n      22.22*m\n---------------------->\n|\n|9.8*m\n|\nV\n\nPeople are most comfortable when they are accelerated into their seats (not the window or ceiling), so the bank angle will be perpendicular to the sum of those two vectors.  Your result should be about 65 degrees (though the 5 isn't a significant digit-- you should be able to get the final answer on your own).\n\nBut oh those poor people!  They'll be accelerated to almost 3 gs!
716722
For launch velocity\nv0 = 4.47 m/s,\nlaunch angle\nθ = 66 degrees,\nThe horizontal range is\nR = 1.515173718712374 m.\nThe total time of flight is\nt = 0.83337718278825 s.\nThe peak height is\nh = 0.8507839727702984 m.
1147435
(tangent 30 degrees= 1km/x) - (tangent 60 degrees= 1km/y)\n\nSolve for the two variables, then subtract y from x.  That is how far the plane traveled in 10 seconds.
1104626
One cannot provide solution without the longitude of local station...\nIt shall be atleast 2500 miles away...
811631
They pass each other at 2:20pm.\n\nOne way to figure this out is by considering the problem on an x-y graph:\n  Think of the x-axis as elapsed time, in hours.  So for example, x=1 would be 1 elapsed hour (or 1:00pm).\n  Think of the y-axis as position, in kilometers.  So think of Austin as being y=0, and Los Angeles as y=2100.\n\nThen, consider the trip of the two planes as two lines (ie. y=mx+b) on this graph.  In the end, youll use these lines to solve the problem.\n\nOk... so the plane that starts in Austin would be y=500x.  The reason we know this is because:\n   1)  at noon (x=0), the plane is in Austin (y=0).\n   2)  if you look at the way we set up the graph, the speed of the plane happens to be the slope of the line (notice "km/hr" = y/x), so the slope is 500.\n\nThe plane that starts in LA is slightly more complicated.  Its y=(-700x)+2800.  We know this because:\n   1)  at 1PM (x=1), the plane is in LA (y=2100)\n   2)  its speed is 700km (but in the opposite direction), so its slope is -700.\n\nSo, we're looking for when the two planes intersect... They intersect when their positions (their y values) are equal.  So you just need to solve the two equations with their y=values as being equal.\n\ny = 500x\ny = (-700x)+2800\n\n500x = (-700x)+2800\nx = 7/3\n\nOk so x = 7/3... Remember that x is elapsed hours, so 7/3 = 2 1/3 elapsed hours = 2 hours and 20 minutes....  or 2:20PM!\n\nYou can verify this by plugging x=7/3 into both of the equations.   You'll get y=1166.667, which puts the planes right around New Mexico =).
763890
You asked this question twice, and I posted a detailed solution in answer to your other question.  I know your teacher's answer is 49 degrees, but if you go to the nearest tenth of a degree, 48.9 is a better answer.
596270
45.987F if u don't believe me - take a thermometer and go around the world
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
241090
it is the period of flying hour times used betwen aircraft servicing. think of it like a car having a 60000 mile service.
72949
8 am to 9 pm Local time.