what was the average velocity in m/s for the whole episode from leaving the nest to returning?
86843
Corresponding to the definition:\n\nav=deltaX/deltaT\n\nWhere av is average velocity.\n\nSo because of deltaX being zero, av is Definitely zero
439730
107.23 mph.  Do I get 10 pts?
950679
(8.5m/s)/.008s = 1062.5 m/s^2 = a\nm = 12\nF=ma=12750N. <----- b)\nImpulse is Fdt, so 12750*.008 = 102 N/sec
986802
91-92 MPH on average. I think he's been clocked at 94 before. And no, it can't go back in time. Without a time machine, anyways.
109202
Fnet= MA so you acceleration is 36km/h, so convert that to m/s, which is 10 m/s. To get acceleration, we know that the car started out at 18km/s, which is 5m/s. So the acceleartion = Change of velocity/Change of time. So that comes out 1 m/s/s. So there is 1000 Newtons on the car. The other guy's answer is wrong. I'm an engineering major. The standard Units in science is MKS, meters kilograms and seconds
837489
neglecting air resistance I hope!\n\ny=1/2at^2+vot+yo\n\na=32 ft/sec^2   you did mean sec^2?\n\nalso, v=at+vo  so we can find vo from this at time 2.1s when v=0\n\nfinally we insert 2.9 s into the first equation (yo= 0 too right?).
1129113
It is 9.8 m/sec , some they have made it 10 :)
544479
The speed of sound: 340.29 m / s
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
103569
Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
95650
It's like the rictor scale (earthquakes).  Warp 2 is twice as fast as warp 1.  Warp 3 is twice as fast as warp 2.\n\nSo say warp 1 goes 1 foot a second. (it goes MUCH faster.)\n\nWarp 2 goes 2 feet a second.\n\nWarp 3 goes 4 feet a second.\n\nWarp 4 goes 16 feet a second.\n\nWarp 5 goes 256 feet a second.\n\nWarp 6 goes 65536 feet a second.\n\nAnd Voyager got to go warp 10 through a poorly written script :)
273335
70 is probably about average.  I always heard it was ideal to get it to around 60 per minute.
577594
12.66 mph\n\nor 12.65558195 to be (more) exact!
340116
a) d= 0.5 * 9.8 * t^2 + Vi * t = 10 m\nFor t=1 sec:\nVi= 10 - 9.8*0.5 = 5.1 m/s \n\nb) g= (Vf-Vi)/ t\nFor t=1sec and Vi=5.1\nVf= 9.8+5.1= 14.9 m/s\n\nHope it is clear enough.
946074
avg. speed = distance/time = 75/11 = 6.82 m/s
932092
Something's wrong with the writeup for the first question.  I'm guessing it should have said THR = RHR + 0.65(MHR-RHR), because otherwise you're gonna get a THR of, like, 10000 beats per minute.\n\nProceeding from that assumption, the THR is RHR + 0.65(MHR-RHR).  You're told that the RHR is 54 bpm.  The MHR is 220 - the age(43), so it's 177.  Therefore the THR is 54 + 0.65(177-54) = 54 + 0.65(123) = 54 + 79.95 = 133.95 bpm.\n\nIn the second question, you didn't provide the "following equations," but think of it this way.  The airplane flew for a total of 15 hours at an average speed of 350 miles per hour.  That means it went 5250 miles.  In the final 7 hours it went 2275 miles.  So in the first 8 hours it had to go 2975 miles.  2975 miles/8 hours = 371.875 mph.  I don't exactly know how to write that in an equation, but now that you understand the reasoning, hopefully you'll be able to recognize the correct equation.  If all else fails, plug the data into each equation and see which one returns 371.875.