If an electric train is going east at 60 miles an hour and there is a strong westerly wind, which way does the smoke from the train drift.
125941
There is no smoke. It's electric.
76048
Modeling the fly and two trains as particles,\nAnd taking the midpoint of trains as origin,\nthe equation of motion of trains are:\nx(t)=-30+18t\ny(t)=30-18t\nin kilometers respectively.\n\nThey collid when 18t=30,\nthat is at t=30/18=5/3 hours (100 minutes).\n\nThe distance traveled by the fly,\nWHICH IS FLYING AT SPEED 80 KMPH,\nin 5/3 hours is\n5 / 3 * 80 = 400/3 km\n\nThat is approximately 133.3 kilometers.
324685
It is easiest to think of this in 2 steps.  First, how long does it take the east bound train going 65 mph to travel the 175 miles to the station?\n\nanswer: 175/65=2.6923 hours\n\nStep 2:  How fast does the westbound train need to go to travel the 63 miles in 2.6923 hours?\n\nanswer 63/2.6923 = 23.4 mph
429887
quantum is wrong, obviously....think about it, if anything travels 35 km, and it takes more than an hour to go that distance, it must therefore be going slower than 35 km/h\n\n35/84 * 60 = 25 km/h upstream\n\n35/50 * 60 =  42 km/h downstream\n\nbecause the current will be adding to the speed downstream, but will also be subtracting from the speed going upstream\n\ncurrent is difference between the two divide by 2 = 8.5 km/h
659238
Because it travels along the least obstructed air.  As air is sometimes denser at certain elevations the lightning flows along the path of least resistence.
596743
The top answer is incorrect.  In both cases, the current woul go on infiniitely, because energy will keep on traveling, but the current will get weaker and weaker, because it will keep spreading aparty, until the ammount of current in negligeble.\n\nHope this helps.
576902
Point A and Point B are 600 miles apart. A train leaves Point A travelling toward Point B at 5:30AM at a constant speed of 65 kph. Another train leaves Point B heading toward Point A at 6:20AM at a constant speed of 40 kph. What time do the two trains meet?1 kilometer = 0.6214 miles\n\nx=kilometers     5/6(50 minutes/60 minutes)*65= 54.166...km\nx=.6214 (600 miles)\nx=372.84 kilometers apart\n5/6(50 minutes/60 minutes)*65= 54.166...km\n372.84-54.17= 318.67km between the 2 trains at 6:20\nthats as far as I can get (I don't remember how to do this... hope this helps!)
98203
I also think that the snow, itself acting as a blanket covering, mutes the noise as the train passes.  Indeed, the corrision issue would likely cause railroads to shy away from such an idea, especially if real water was involved, which is very damaging to the tracks.
514166
Southbound: 150 km/h\nNorthbound: 200 km/h\n\n\nThey're moving apart at 850 - 500 = 350 kph.\n\nSo at 9 am they were 500 - 350 = 150 km apart.\n\nWhich means that South was going at 150 kph;\n\nSo North was moving at 200 kph.
561709
It's all relative. Easiest way to see the answer to this problem is to first imagine the river and log stationary. Assuming the boat keeps the same speed S, he travels an hour and then turns around and takes an hour to get back(a total of 2hrs). If the river/log has some speed v then when he travels away from the log he has speed S+v (with current) but since the log has speed v he travels at speed S away from the log (same as the stationary case) and on the way back has speed S-v (against current) and the log has speed v  and so he travels at speed v+S-v = S towards the log. So it is the same as the stationary case: one hr out and one hr back = 2hr.
167412
well if the train is a mile long and the tunel is a mile long then as soon as it hit the tunnel it would have to travel 2 miles to get the whole train through the whole tunnel so since he travels 60mph then it's a mile a min. therefore he needs to to travel 2 miles so it would take him 2 minutes to get through it.
537573
Lots of unnecessary complications here.\n\nAirspeed is the hypotenuse. 20.4 is the side opposite the angle we are looking for.\n\nThe heading, east of north, must be at angle sin^-1(20.4/241).\n\nSpeed relative to the ground is the third side of the triangle, and is \n\nsqrt(241^2 - 20.4^2)\n\nZidane has got the angles all wrong. The wind is from the east. We want to fly due north. That's where the right angle is. 241 is the hypotenuse of the triangle. The wind serves to make the ground speed less than 241, not greater.
611504
I would say on the eastern side.  You're still going to be experiencing the most Positive Vorticity on that side.  Though the clouds should be mostly mid and high level clouds.  \n\nAs for formation:\n\nFrom the American Meterological Society glossary:\nA sequence of events by which a warm high or cold low, originally within the westerlies, becomes displaced either poleward (cut-off high) or equatorward (cut-off low) out of the westerly current. \nThis process is evident at very high levels in the atmosphere; it frequently produces, or is part of the production of, a blocking situation.
605417
It will be more. Mathematically:\n\nno wind case:\n\nLet the distance from A to B = d\nAnd let the speed of the plane without wind = v\nSince distance/speed = time, in the case with no wind the time for the return trip =  d/v + d/v = 2d/v\n\nWind case:\n\nLet d,v be as above, and let the wind speed blowing from A to B = w. The planes wind-aided speed = v+w and it's wind reduced speed is v-w. The time for the return trip is now:\n\nd /(v+w) + d /(v-w) = (d(v-w)+d(v+w))/(v^2 - w^2)\n\nThis simplifies to:\n\n2dv /(v^2 - w^2) compared to 2d /v for the no wind case.\n\nThe ratio of the no wind case to the wind case is:\n\nv^2 / (v^2 - w^2)\n\nWe see that if w > 0, this ratio will be greater than 1, which means the return trip will take more time if there is a wind.
337167
Well, when you jump out of a moving train, you will be carried forward in the direction of its movement, this is because when the train is travelling, the train is moving with a velocity of v towards the x-component. Therefore, as you are on the train, you are also moving with the velocity of v towards the x-component. Hence, during your jump, you are still 'travelling at the x-component' with the velocity of v. So, you will be carried forward. Hopefully I answered your question and make you understand better.
17169
In the back. \n\nThe train has just left the station & is still accelerating as it is leaving the station. So, the train will be moving faster when the back of the train enters the tunnel & the claustrophobic will spend less time in the tunnel.
157106
Only the wind is dependent on the radiant energy from the sun.  Wind is caused by uneven heating on the Earth's surface.  That heating is caused by the Sun.\n\nFossils fuels need air to burn and are from the ground\nHydroelectricity is from flowing water--depends on gravity.\n"All of these" can't be a choice if at least one can be eliminated from the list <-- a multiple choice test rule!\n\nThis is probably a test question.  The operative word in this question is "dependent".  Take it from a retired New York State Regents Earth Science teacher.