A bus moving with speed 0.99c is 80 feet long. What is its length your vantage point on a fixed planet?
441829
Its length?\n\n9 inches baby.\n\n;)
311411
Math model of problem:\n\nD2S/Dt^2 = 2.1  \n\nIntegrating:\n\nDS/Dt = 2.1*t + c1\n\nIntegrating again:\n\nS = 2.1*t^2/2 + c1*t + c2\n\nFrom the statement of the problem:\nWhen t=0 DS/Dt=0 and S=0\nTherefore c1 and c2 both are zero\n\nSo S = 1.05*t^2\nSo t = square root of S/1.05\nThat is t = square root of 70/1.05\nthus t = 8.165 secs.\n\nSo boat accelerates for 8.165 secs in travelling from rest over a distance of 70 metres
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
230260
First, the Earth is rotating at about 1,600 km an hour. Second, a full Earth orbit takes 365.242 Earth days, or about 8,766 hours. We can find the "circumference" of Earth's orbit by taking the radius (about 150,000,000 km) and multiply by 2pi, giving us about 940,000,000 km. Rate equals distance over time, so (940,000,000 km)/(8,766 hours) equals about 107,300 km per hour. \n \nSo far: In only the solar system, Earth's maximum speed is about 109,000 kmph and minimum speed is about 105,500 kmph.\n\nHowever, the solar system is orbiting around the center of the Milky Way galaxy. We are about 2/3 of the way out on the 100,000 light-year disc, or about 624,394,875,000,000,000 km. It takes us 200 million years to orbit. I'll shorten this and not completely lay out the math..................\n\nroughly 2,240,000 kmph.\n\nAdd this with our previous number to get:\n-Earth's maximum speed: roughly 2,357,000 kmph\n-Earth's minimum speed: roughly 2,130,000 kmph\n\nBut wait: we're not done yet. The Milky Way is speeding through space towards the (relatively) near-by Andromeda Galaxy at 600km every second, or 2,160,000 kmph. Factor this in to get our final result:\n\n-Earth's maximum speed: roughly 4,500,000 kmph\n-Earth's minimum speed: roughly 197,000 kmph\n\n\nSo how does all of this confusing math relate to your question? Look at this set of data. Regardless of the Earth, the Milky Way is cruising along at roughly 2,160,000 kmph. The Earth, however, relative to the Milky Way, is moving as much as about 2,357,000 kmph and as little as about 2,130,000 kmph.\n\nSorry this is so long winded.
256497
around 1/6th that of earth so around 1.64m/s
849468
gravity at the earth's surface = 9.80 m/s^2
741279
Total distance traveled divided by total time in transit.\n\nr = d/t.
671105
almost exactly, assuming someone on earth is doing the measurement. someone moving very fast would get a different result.
996962
The speed of light is a constant (186,000 mi per second) and does not change based on what it is travelling through.
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.
871085
let me go check it, 924,375,700 km\n\n:> peace\n.
703424
This is a very basic question and you need to get extra help if you are having trouble solving this in your physics class.\n\nAcceleration is force / mass.  (from F = ma).\n\nThus, for the astronaut\n\n120 = 79(a)\na = ~1.52m^2/s\n\nAnd the satellite\n\na = 0.182\n\nNow they travel in opposite directions (BOTH of them move). So calculate how far the astronaut travels, how far the satellite travels, and add them. Simple kinematics:\n\np = 2a(t^2) + v(t)
471913
The daily path of an object on the celestial sphere, including the possible part below the horizon, has a length proportional to the cosine of the declination. Thus the speed of the diurnal motion of a celestial object is this cosine times 15 °/hr = 15'/min = 15"/s, i.e. (compare angular diameter):\n\nup to a Sun or Moon diameter every two minutes \nca. four seconds for the largest planet \n2000 diameters of the largest stars per second
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.
32272
well Bus B travels less distance in the same time, meaning the velocity is greater for bus B and since velocity is the slope on a distance time graph, the slope of the graph would be steeper for bus B than for bus A.\n\nif you need a numerical answer, the slope for bus A would be:\nv= d/t= 300/12= 25 m/s\nand the slope for bus B would be:\nv= d/t= 200/12= 16.67 m/s
448267
the average orbital velocity of earth is 29.783 km/s.\n\nhttp://en.wikipedia.org/wiki/Earth
792509
At 1 p.m the 1st bus starts moving west with a velocity of 44 mph  that means at 2 pm the bus will be 44 miles away from the station because  D = V x t = 44 mph x 1 h = 44 miles .\nAt 2 pm the 2nd bus starts moving east i.e in the opposite direction with a velocity of 48 mph \n              \nB1  ____________B2      ( B1 B2 = 44 miles )\n                      \n             \n\nAfter 2 pm the distance between the 2 buses at any time is equal \nto : \nD = 44 +( V1 x t ) + (V2 x t ) = 44 + (44 x t ) + (48 x t ) = 44 + 92 t \nwhere D >> miles , t >> hrs \n                                         \n                                          B1    _________S___________ B2\nWhere B1 S = 44 miles + 44 t \n            B2 S = 48 t \n                                             \nPut D = 274 miles to get  t \n274 = 44 + 92 t     >>  t = 2.5 h\nTherefore the two buses will be a distance of 274 miles apart at : \n2 pm + 2.5 h =  4 : 30 pm