At 5 p.m., Roger whizzed off from Asteroid. One hour later, Maryanne whizzed off from Asteroid in the opposite direction at a speed 400 kph less than that of Roger. If they were 7900 km apart at 11 p.m., how fast did each travel\n\nAnswer in the form: Roger, Maryanne
209347
Roger - 900kph\nMaryanne - 500kph
330981
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
494173
Both balls would hit the rocks at the same speed.\nBall #1 would leave your hand travelling straight up at speed 'u'. It travels up, stops, and starts plummeting back to earth. When it reaches the height at which it left your hand, it is travelling at speed 'u' straight down. The speed that it hits the rocks at is u+(time x 10m/s/s).\n\nBall #2 would leave your hand travelling straight down at speed 'u'. The speed that it hits the rocks at is u+(time x 10m/s/s). It is the same speed.
656971
WOULD SOMEONE PLEASE EXPLAIN RELATIVITY TO THIS GUY I'M ALREADY 30 POINTS  DOWN ON THE DAY AND ANN FURROWS IS LANDING ON KZORB AS WE SPEAK
300228
V = (v + u) / (1 + uv/c²)\n\nTry that. The velocities are quite high so relativistic. Haven't tried it yet though.\n\nNevermind, the velocites aren't nearly high enough to be relativistic. If it was km/s it would work.
511220
1) The total rocket, with mass 3M travels 260m over the first "half" of the parabola. On exploding, the fragment with mass M falls straight down - i.e., all the momentum in the horizontal direction remains with the larger fragment. Therefore its speed in the horizontal direction will increase:\n\nP = constant\nBefore explosion, P = (3M)(u)\nAfter explosion, P = (2M)(v)\n\nTherefore the speed in the horizontal direction will increase to 3/2 times the original speed.\n\nThe vertical momentum of the larger fragment will not increase, as the smaller fragment falls as from rest i.e. no original momentum. Therefore the time taken for the larger fragment (indeed, either fragment) to reach the ground from the apex of the motion will be the same as it was to reach the apex from the starting point. Since the horizontal distance was 260m, the extra distance covered will be that multiplied by 3/2, since the speed increased by that much, so the total distance from the launch point is 260 + 3/2(260) = 650m.\n\nYou do it by considering the horizontal and vertical motions independently.\n\n2) If the bobs collapse elastically, then no energy is lost in the collision. The energy of the larger bob on contact is entirely its kinetic energy - this comes entirely from the potential energy it had at the top of its swing. If the angle is 15deg and the length of the cord is 50cm, then the bob is drawn back until it is at a height of 1.7cm. The formula for potential energy is U = mgh, so the potential energy of the bob at that point is \n\nU1 = (0.39kg)(9.8m/s2)(0.017m) = 0.06497 joules\n\nconverting all units to SI.\nThis is the kinetic energy of the bob as it strikes the second one - all of this energy is transferred to the second bob, as kinetic - the second bob rises until it stops, when all the kinetic energy is converted to potential - and so use the same formula to get the height it reached.\n\nU2 = 0.06497 J = (0.14kg)(9.8m/s2)(H)\n\nwhere H is the maximum height of the lighter bob. It works out as\n\nH = 0.047 m\n\nor 4.7 cm. Using simple trigonometry, this corresponds to an angle of 25.1 deg.\n\nHope this helps
524273
Well. \n\nHow fast does sound travel in three seconds?\n\n3x 340 m/s = 1020 m. I thought that was \na little fast, but I could be wrong.\n\nLight travels way faster than that, so you pretty\nmuch see the lightning right when it happens.\n\nSo the event occurs 1020 meters away.
117916
ok well his initial velocity is 0 and the distance is 402m and time is 4.913. you have to find the acceleration. well the only way i can think of doing this  without using these basic equations that are derived from newtons law (i forgot the derivation) is this.     you find the average velocity needed to go 402m in 4.913s. which is just 402m/4.913s = 81.823....\nthis is just the average velocity and because acceleration is constant, this means you can just find the average by doing  Vi + Vf / 2. you have Vi =0    so Vf/2 = 81.823........ so Vf = 163.6474.........m/s\nacceleration is  change in velocity /time. you have change in velocity  (Vf-Vi)   so divide it by time.   \n163.6474-0/4.913   =  33.3 m/s^2\n\n2. so the acceleration due to gravity is 9.8m/s^2. you know the distance is 10m and initial velocity is 0. so Vf = a x t\nso Vf = 9.8t. and Vf+Vi/2 = d/t    same things basically. so Vf/2 = 10/t  which means  Vf x t = 20\nVf = 9.8t\nVf x t = 20\nso then that means that (9.8t) x t = 20\n 9.8t^2 = 20\nt^2 = 2.04....\nt = 1.4285.....  so it takes her that long\nand now you just find Vf by putting this time in the equation  Vf = 9.8t = 14m/s. thats final velocity\n\nb. ok so this question confused me at first but it snot hard. once she hits the water (with speed of 14m/s), she travels for 2m before coming to a stop. so distance is 2m. Vf = 0. Vi = 14m/s and you have to find acceleration. \nso a = (Vf-Vi) x t = -14t\nso you need time now. you do average velocity  again\nVi + Vf / 2 = d/t\n14/2 = 2/t\n7 = 2/t\nt = 0.2857.... s\nthen put it in the other equation and you get a = -14t\nso a = -4m/s^2.\nforce = mass x acceleration\nmass is 61kg so you do 61kg x -4 = -244 N\nthe water exerted 244N against the movement (thts why its negative).\n\nso hope this helped. let me know if you need more help or whatever. laterz
102107
3350 + 100t = 4850 - 150t\nWhere t is time in hours\nrearranging\n4850 - 3350 = 100t + 150t\n1500 = 250t\nt = 1500/250 = 6\nthey meet in 6 hours\n3350 + 100*6 = 3950\n4850 - 150*6 = 3950 wahoo!\nNow the only question is "do they survive the impact?"
125717
OK, first lets say that the rocket moves at a constant speed of 80% of the speed of light. Working with an accelerating rocket would be hard and .8 is a convienent number.\n\nso we have beta=.8 \ngamma=1/sqrt(1-.8^2)=5/3 and \nbeta*gamma= 4/3\n\nIf you dont understand this, look up lorentz transform in your textbook.\n\nWe will use a two dimensional unprimed coordinate system with distance in light minutes on the x axis and time in minutes on the y axis. The sun is located at x=0 and the earth at x=8 light minutes. Draw 45 degree lines, representing light rays, intersecting the x=0 axis at one minute intervals. Draw a vertical line at x=8 representing the earth. The light ray starting at (0,-8) intersects the earth at (8,0) meaning this is the event where people on earth observe what happened on the sun 8 minutes ago. We will call this event A.\n\nDraw a line with a slope -1/.8 intersecting A. This is the path of the spaceship. It will reach the sun at the time 10 minutes.\n\nNow we will calculate the cooridinates of event A as seen by the spaceship approaching the sun. We have\n\nA'= [gamma       beta*gamma]  [8]\n.......[beta*gamma  gamma     ]  [0]\n\nA'= 1/3 [5 4] [8] =  [13.3]\n.............[4 5] [0]  ... [10.6]\n\n(hopefully the ... makes the matricies line up)\nSo the guy in the spaceship NAMES this point 13.3 light minutes and 10.6 minutes.\n\n\nNow lets look at where the line of the light ray from (0,0) intersects the path of the spaceship. Using trig we get (4.44,4.44). Call this point B. This is where the guy in the spaceship observes what happened on the sun at time 0.\n\nNow lets lorentz transform this point\nB'= 1/3 [5 4] [4.44]  = [13.3]\n.............[4 5] [4.44] ...   [13.3]\n\nnotice that the x' coordinate is 13.3 for both point A' and B'. This is because from the perspective of the guy in the ship, he has not moved. The elapsed time for the guy in the ship between A' and B' is 13.3-10.6 = 2.7 minutes.\n\nSo, he has seen 8 minutes worth of dancing in just 2.7 minutes. The image has speeded up.\n  \n* This is the correct method for solving this problem, but you might want to double check my math.
29731
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
399774
So(a). Initial kin energy E=m*V^2/2, m=100kg, V initial speed to be found.\nElementary work of gravity dw=F(r)*dr, where F(r)=G*M*m/r^2 is gravity force at height r considered from center. Thus W=Integral {for r=RE until 2*RE} of G*M*m*dr/r^2 or W=G*M*m/(2*RE); \nAs m was launched right under vertex, then E=W\n or V=sqrt(G*M/RE).\n\nAnd(b). if m is placed somehow in the orbit, then centripetal force f=m*v^2/RE, which = to G*M*m/RE^2, its kin energy E=m*V^2/2 or E=0.5*G*M*m/RE is work in question.\nIn real life all this stuff is more bizarre and sophisticated.
568549
so do they start from the same place? \n             rate    x     time            =       distance\nJack  5 (mph)      x (hours)                  5x (miles)\nJill      8 (mph)   x-.5 (hours)               8x-4 (miles)\nthe distance when jill catches up with jack will be the same since they have both traveled the same distance\n5x=8x-4\n4=3x\nx=1+1/3\nso jacks been going for 1+1/3 hours or 1:20. that much past 1:00 pm is 2:20\nI hope this makes sense to your wife!
320123
The difference in rates is 1 km/hr. So if they are moving in the same direction, the faster walker will be 5 km ahead at the end of 5 hours.\n\nThe sum of the distances traveled per hour is 11 km. So at the end of 5 hours of travel in opposite directions, they will be 55 km apart.
485020
Hmm-m. Interesting question...\n\nI thought about this for a while and see no\nreason why this math would not be correct.\nAssuming that the bullet did explode when\nstruck by the firing pin, and the bullet left \nthe end of the barrel of the gun at say \n2000 feet per second velocity...\n\n2000 fps = 1800 Mph\n\nThe Moon is 239,000 miles from Earth.\nLet's say you were 10 miles up in the sky\nwhen you fired the gun.  Distance = 238,990 Miles.\n\nSo, 238,990 / 1800 = 132.7 hours \n\nSince there is no atmosphere in space to slow down the \nbullet as it traveled, we can estimate that the bullet would \ncontinue traveling at 1800 mph for the entire trip. The \ngravitational pull of the Moon would influence the bullet's\nvelocity somewhat towards the end of its flight, but my\nfigures show that with 1/6 the gravity of Earth, the Moon\nwould not significantly change the time of flight.\n\nAnswer: 132 Hours.   or   5.5 Days.\n\nNow, I have two questions for you:\n\n#1.)  If the rifle did fire off in space, would you see any smoke?\n\n#2.)  If you were 10 miles up in the atmosphere - say 50,000 ft.,\nhow fast must you be going, to be able to remain at that altitude,\nand in which direction?\n\nYour question makes reference to the Space Station as the point of firing the gun. In my answer above I did not use that as the platform from which the gun was fired because I do not have at my finger tips the approximate altitude of the Space Station though I know it is much more than 50,000 Ft. I also did not take into account the thin atmosphere which exists at 50,000 feet, or the pull backwards on the bullet by the gravity of Earth. These omissions were made simply to run the numbers and see what I came up with since I thought this was an interesting question. Now, having dispensed with the mundane aspects of my quick math, let's discuss the real problem with this question...\n\nThe Earth is spinning. The Moon is orbiting the Earth at a certain velocity, hence it is a moving target. The moving target is being shot at from a moving platform. If you have any hope of hitting the Moon with your bullet, you would have to fire at the place where the Moon will be 5.5 days from the time of firing your gun, not at where the Moon is now. To do that requires more math than a few simple divisions of distance and velocity. For example, the moon swings completely around the Earth at a radius of 238,900 miles in 27.3 Earth days. The circumfrence of that circle is Pi x D = 1,500,292 miles (assuming the orbit is a circle, which it is not). So, the velocity that the Moon is moving away from its present spot right this minute is 2,290 miles per hour.  Since the Moon rises and sets on our horizon each night, you can see that in our calculation, the Moon would have passed overhead at least 5 times in 5.5 days. So your rifle would have to be pointed at the place you calculate the Moon should be at in 5.5 days.        \nRegards,\nZah
511629
K so...I haven't done any Grav froces recently...but I'll try\n\nIm guessing use  Net force = ma\nThe net force is the Gravitational force, Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2\n\nfor the 4 (pie)^2 r/ T^2    r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.\n\nand the other r is the planet's radius\n\nm2 cancels\n\nsolve for m1, and yo uhave the mass of the planet.\n\nI guess plug in numbers after that, into the equaltion Fg = Gm1 m2  /  r^2\n\nm2 would be the 70 kg\n\nhope thats right
519771
Of cause, help.\nthe force of gravity is mM/R^2 = 550 N\nThis is for R = 3400km\n\nIn case we take the object up for 12 km, the distance from the centre of the planet will be 3412 N.\n\nand its weight will be mM/(R+12km)^2 = Wn\nWe can find  Wn   just from these two equations:\nfrom the first equation we shall express \nmM = 550(N)xR^2\nthan  Wn = 550 (N) x R^2/(R+12)^2 = 550(N)x3400^2/3412^2 = \n550 (N) x 0.996483^2 = 550(N)x0.9929783 = 546.138 N\nThat is the answer