I thought 64/m^5/9 am i right or wrong please explain bc i do not understand this question.
120513
9 - 48m^2 or 9(1-4m^2)\nthe 1 before the second set can be dropped, it's implied. Read it like this:\n(3*3)-(3(4m*4m))\n(9) - (3*16m^2)\n9-48m^2\nThen divide it all by the common 9\n9(1-4m^2) : That's your answer
450458
You have mixed units. To solve, units must be consistent. Speed is in miles per hour so time must be in hours.\n\n210=(70/60)(y-x)\n210=(50/60)(y+x)\n\n1260=7y - 7x\n1260=5y + 5x\n\n6300=35y - 35x\n8820=35y + 35x\n\n2520=70y\n\ny = 36
402576
try: \n\nF=m*a =>     0.3*M*g = M*a    =>    a=0.3*g\n\na*t = 60 miles/hr   =>    \nt = 60miles/hr / (0.3 g) = (60 miles per hour) / (0.3 * 9.8 ((m / s) / s)) = 9.12 seconds\n\ndistance\nD=1/2*a*t^2 = ((0.5 * 0.3 * 9.8 * (m / s)) / s) * ((9.12 sec)^2) = 122.266368 meters
630083
Jennifer, I'm a little bit puzzled about your question because you're asking for what the funtion is and you have it written there.  The funtion is s = -16t2 + v0t + s0\n     What I really think you want is the solution.  Correct?  Well, here's how you get that.  You put in 0 for s0, and 64 for v0.  ALSO, s, on the left side of the equation, is also zero because it will land back on the ground when it's done.  When you solve using these numbers, the only variable left is t.  And when you do the algebra, \n  0  = -16t2  +  64t  + 0\n \n16t2  =  64t\n\n16t  =  64\n\nt  = 4      seconds, that is.\n\nI hope this works.  I'm a physics teacher.  If you need further clarification for this problem, or if you need help with any others, feel free to write me at fortitudinousskeptic@yahoo.com\nI'll be glad to help!
190589
I picture an axle with one wheel bigger than the other for this problem. Since they're not spinning, wheel A has to cover the same amount of distance wheel B covers in the same amount of time.\n\nWheel A has radius 12cm and rotates at 35 radians/second:\n12 cm * 35 rad/sec = 420 cm / sec.\n\nWheel B must also cover 420 cm each second. So we work backwards:\n\n(420 cm / sec) / 17 cm = 24.7 rad / sec\n\nSo you're right, B is the correct answer.
495544
First problem: You have assumed all the mass of the meter stick is concentrated at the lower end; obviously, it is uniformly distributed along the length of the stick.  Assume that the mass is concentrated at the center; then, the length of the oscillator will be 0.865 meters, and the period should correspond.  There are some complications involving moment of inertia which I am ignoring.\nSecond problem: You may be solving for the wrong thing.  It looks like you are solving for the frequency of oscillation, rather than the maximum speed of motion.  Some careful work with units will help show you what is going on.
428762
well F=mv/t given: m=78kg, v=24m/s, t=60s\n\nF=78(24)/60=31.2N
489517
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
29156
That is right. Just put them in a calculater to check. 2/9=0.22222222 and 8/36=0.2222222. Either they are wrong or you misinterpred the question and it was asking something else. The best way to solve a problem like that is to cross multiply. 2/9=8/? turns into 2*?=8*9.
551385
Gopal has the correct answer! \n\nSo... becareful with some other results... and... never use:\n0.833 instead of 5/6... it's a mistake!\n\nFinished Algebra with 19!
357063
ya, that one has the same dimension as that of time.see how that eqn is derived.you find distance covered by 'n' seconds then distance covered by (n-1) seconds.subtract.and you get the eqn.... but see wat is the difference b/w the time n and n-1. it is one second..this is the one that appeared in the eqn and confused you.is it clear?
153796
M = Martina's distance from home at time t, in hour.\nM = 24t\n\nJ = John's distance from home at time t, in hour.\nJ = 48(t - 2) \n\nWhy (t - 2)? because, John started 2 hours later.\n\nThus, when " john catch up with Martina" means that their distance from home is equal. Thus, \n\n24t = 48(t - 2)\n  t = 2t - 4\nor t = 4.\n\nThis means 4 hours from the time Martina's leaves home, John, would have caught up with her. The time is just 4 hours after 9 am, i.e, 1 pm.\n\nRegards
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
413210
Your answer key is wrong, or you are not reading the right answer.\nFrom the data you presented above y = x^3 is OK.
612840
First of all, lets just look at 12 hrs, as it will be the same for the next twelve:\n\nThe minute hand revolves around the clock at 1 rev per 60 minutes, the hour hand revolves around at 1 rev per 60•12 minutes.  Knowing that after 60 minutes the minute hand is back to where it started we can say that 2π≡0.  We want to rewrite the above relations to give us the positions as a value of time.  This can be done by assuming that t is time (in minutes) and using m=2π/60t (minute hand) and h=2π/(12•60)t (hour hand).  \nTherefore we want to solve m=h+2πk for some integer 0≤k≤10.  The reason that we don't have k go to 11 is because if you think about it, the hour and minute hand never cross during the 11th hour, they keep getting closer, but then at 12:00 (during the 12th hour) they meet.\n\nThus 2π/60t=2π/(12•60)t+2πk ==>> 12t=t+720k ==>>11t=720k ==>> t=720/11k.\n\nNow, this is the time after 12:00 (midnight or noon, I don't care they; are the same). \nThese times are:\n\n12:00 (k=0)\n1:60/11=1:05:300/11≈1:05:27.27 (k=1)\n2:10:600/11≈2:10:54.55\n3:16:240/11≈3:16:21.81\n4:21:540/11≈4:21:49.09\n5:27:180/11≈5:27:16.36\n6:32:480/11≈6:32:43.64\n7:38:120/11≈7:38:10.91\n8:43:420/11≈8:43:38.18\n9:49:60/11≈9:49:05.45\n10:54:360/11≈10:54:32.73\n\nI decided to give you the second values as fractions also, since they can't be given accurately by a terminating decimal.
77544
Let x be the side of a square\nThen, area of the square is, A=x^2\nThe rate of change of area w.r.t side is dA/dx =2x\nIn the problem another parameter is given, that is time.It is given in the problem that there is a constant change in the length of the side at the rate 0.4cm /sec.Here we have to notice that corresponding change in area is not uniform throughout.Therefore we can only find instantaneous change in area or accumulated area in the given periodor interval.That means change in area at a given instant of time or interval.\nFor example, let initial area be 0.Now, to find the area between the interval a and b we have to reintegrate. ie. integration of dA between the interval a and b =integration of 2x in the interval a & b.\nie. [A]a&b= [x^2]a&b=b^2-a^2 square units\nFor instance, change in area between the interval 9th and 10th seconds is =[x^2]a&b=b^2-a^2 square units\nGiven, 0.4units/sec,considering intial length and time as 0,we have, length of the side on the 9th second is a=9*0.4=3.6units and in the same way length on the 10th second is b= 4units\nAccordingly change in area(accumulated area) at the 10th second is =b^2-a^2\n=16-12.96=3.04 square units [this is an instataneous change]\nPlease note that, this change cannot be written as units^2 /sec
539466
rate b = rate a / sqrt (mwtb/mwta)\n          = [rate a * sqrt mwta]/sqrt mwtb\n          = 0.0235 * sqrt64 / sqrt 352\n          = 0.188/18.7624 = 0.01 = 10 x 10^-3...aprrox the same answer\n she got that 9.88 by taking m wt UF6  as 362....lol\nplug it in to see...\nbtw, your algebra was wrong....