80kg mass of 10kg speed of 2.0m/s?
395993
What kind of statement (question) is this?
210248
Top Speed 330.0 kph / 205.1 mph \n0 - 62 mph 3.8 seconds
788979
For the add on: If a toy plane weighs two pounds, how fast should it accelerate? What unit of measurement would you use?\n\nNot sure what you're asking. Unless you're talking about how fast it would accelerate if you drop it. Then it would be vertically down at 32 ft/s^2 in a vacuum. No matter how much it weighs. (I'm using the English system units since you talked about pounds.)\n\nIf you're asking for an opinion of what would be a nice number for acceleration ... Well, if you wanted it to get up to a speed of 50 ft/s within 20 seconds:\nV = a*t or acceleration = V/t = (50 ft/s)/20 s = 2.5 ft/s^2\n\nI hope this helps
618887
By relativity, E = Mc2\n\nIf E = 4 X 10^6 J, then....\n\n4 X 10^6 / c^2 = 4.45 X 10^ -11 kg\n\nWhere C = 299,792,458 m/s
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
909202
a= r * w^2 = .12 * 2^2 = .12 *4 = .48 m/s^2
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
30694
20m/s.\nbecause the total time is 5s. and by defn of average speed it is the ratio of total distance travelled to total time taken i.e 100/5 = 20m/s
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
537369
If your mass is m, multiply it by g to convert it to N, where, g is the acceleration due to gravity of the place where you currently are.\nFor most of the places on earth, we take the average value of g = 9.81 (m/s)/s.
695486
I hope I still have the equations memorized. I will try to give you a helping hand, but please check the answers.\n\nMomentum is = mv.... mass times velocity\n\nM= mv = 1.24*10^5 * 75.5 meter per second = 9362000 kg.m/sec\n\nHorizontal component of the momentum is equal to Moment cos(x), where x is the angle of approach which is 11.5. The equations are:\n\nVert component is = mv*sin(11.5)=1,797,504 kg.meter/second\n \nHoriz component = mv cos(x) = 9,174,760 kg.meter/second
754995
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5
1035224
v^2 = 2gh\n\nh=1/2 gt^2\n\nsubs this.\n v^2 = 2g (1/2 gt^2 )\nv^2 = 2 g^2 6^2\nv^2 =  100 X 36\n\nv= 60m/s
371210
3 * 10^8 m/s (approx.- this is used for the purpose of calculations)\nthe exact value is hard to calculate but the closest approximation is 299,792,458 m/s
969410
Given: m=1.6*10^7kg\n           F=8.5*10^5N\n\nI only guessed that the above is what you meant.\n\nSolve for a, using the formula:\n\nF=Ma\n\n8.5*10^5=1.6*10^7a\na=8.5*10^5/1.6*10^7\n  =5.31*10^-2m/s^2\n\nv=70km/h =70*1000/3600m/s\n  =19.44m/s\nu=0\n\nTo solve for time, t, use the formula:\n\nv=u+at\n\n19.44=0+5.31*10^-2t\nt=19.44/5.31*10^-2\n =3.66*10^2\n =366s
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km