finding the speed of an object...with the mass and the height of sometihng?
816057
again ... conservation of (mechanical only here) energy\n\npot energy = kin energy\n\nmgh = 1/2 m V^2\n\nso V = sqrt(2gh) -- see mass irrelevant\n\nV = sqrt(2*9.8*8.53) = 12.9 m/s\n\ng=9.8m/s^2 is the acceleration due to gravity\n\ng=10m/s^2 is not a bad approximation
844533
from what I remember conservation of momentum states that: \nm1v1=m2v2\nwhere in this case, m1= mass of the astronaut\nm2=mass of the sat.\nv1= unknown\nv2=speed of the sat.\n\nyou can find the speed by solving the above equation for v1:\nv1=m2v2/m1\n\nFrom there, once you have the speed you can calculate the distance by multiplying the speed (v1) times the time (7.5 s) given in the problem.
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
910291
As long as you know mass and Force, you can solve for acceleration with the equation F = m a \n\nIf you are already learning kinetics, I'm sure you know how to solve for "a". \n\ngood luck.
815192
d = 1/2 a * t^2\n98m =(1/2) * 9.8m/s/s * t^2\n20 sec^2 = t^2\nt = 4.47 sec\n\nd = v * t\nd = 27m/s * 3.16sec\nd = 120.7 m
851569
final speed=initial speed+(acceleration*time)\n=10+(30*30/60)\n=10+15\n=25km/h\nhope that helps..\ngood luck
531134
Depending on the resources you have available you could do an experiement with rockets (small ones foudn at the store). With this you could measure the time in the air, distance traveled, speed would be difficult but you could get an average if you had the distance and time. And by changing the wieght/structure of the rockets would give you measurements on mass. You would have to keep the angle at which you launched them at a constant. (keep in mind the wind factor if out doors)\n\nOther wise if you are looking to only hit one of those topics I suggest testing the mass of objects using water displacement.
1069779
Draw a free-body diagram of the ball and examine it's accelerations - if you assume wind and air resistance are negligible and that the ball is accelerated straight up, then you're left with its initial acceleration (68.57 m/s^2) and gravity (9.8 m/s^2 if I remember correctly) \n\nThe ball will decelerate by 9.8 m/s every second, you can divide to find the amount of time the ball takes to reach its apogee (point of zero acceleration), then integrate the acceleration curve to find the velocity curve, which can be integrated again to find the distance traveled.  I'm not sure it can be done without the calculus.
813425
F = GmM/r^2; where G is a constant, m and M are the two masses, and r is the distance between them.\n\nm + M = 4 kg; so m = 4 - M\n\nF = G(4 - M)M/(.25)^2 = 2.5*10^-10; where G = 6.67 × 10−11 N m^2 kg^−2 \n\nUse the above to solve for M, then m = 4 - M to find m.  \n\nAnd there you have it, you do the math.
587714
First Question:\n\na) Solving for the height of the 1st brick\n\nh=1.00m (sin 30.0)=0.50 m\n\nSolving for the velocity of the 1st brick at the bottom\n\nKE at the bottom = PE at the top\n1/2 v^2 = (weight) h\n1/2 v^2 = 24.3 (0.50)\nv = 4.93 m/s\n\nb) momentum before = momentum after; momentum is just m*v\n\n(24.3/9.8)(4.93)+(35.7/9.8)(0)\n=[(24.3/9.8)+(35.7/9.8)] [v]\n\nv = 2.0 m/s\n\nc) F = ma\n-5.0 = [(24.3/9.8)+(35.7/9.8)] a; F is opposite to motion\na = -0.817 m/s^2; deceleration\n\na =(vf-vi)/t; vf=final velocity; vi=initial velocity\n-0.817 = (0-2.0)/t\nt = 2.45 seconds\n\nd) 2ad = vf^2 - vi^2\n2(-0.817)s = 0^2 - 2^2\nd = 2.45 meters\n\nSecond Question\n\na)Make West East axis as the x-axis; N S axis as y-axis.\n\nx momentum before = x momentum after\n\n880(0)+1585(12) = (880+1585)(vx)\n\nvx = 7.72 m/s; x component of final velocity\n\ny momentum before = y momentum after\n\n880(-16)+1585(0) = (880+1585)(vy); compact car is headed in the negative y-axis.\n\nvy = -5.71 m/s\n\nv^2 = vx^2 + vy^2\n\nv^2 = 7.72^2 + (-5.71)^2\n\nv = 9.6 m/s answer\n\nangle = Arctan 5.71/7.72\n\nangle = 36.5 degress south of east\n\nb) F = coef of friction (weight of the cars)\n\nF = 0.58 (880+1585)(9.8)\n\nF = 14011 N\n\nF = ma\n-14011 = (880+1585) a; - because opposite motion\n\na = -5.68 m/s^2; decelerating (-)\n\n2ad = vf^2 - vi^2\n\n2(-5.68)d = 0^2 - 9.6^2\n\nd = 8.11 meter from the poin of impact
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
153995
F=m*a\na=F/m = 3.5 x 10^7N/(2.0x10^6kg) = 17.5 m/s^2 or 1.8g
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
204311
It's around 23.6 mph (penny falling flat) to 73.3 mph (penny falling on its side).\n\nVterm =  sqrt ((2 * m * g) / (c * p * A))\n \nwhere:\nm=mass of the object \ng=acceleration from gravity \nc=coefficient of drag \np=density of air \nA=Area of the object \n\nIt's pretty straightforward.  The full explanation is at the source.
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
80396
Yeah i know:\nDo this:\n\na.)Find the work done by the applied force.\n\nb.)Find the work done by friction\n\nc.)Find the work done by gravity\n\nd.)Find the work done by the normal force.\n\ne.) IF the object starts at rest, how fast is it going after the 5.0m.\n\nhope this helped\n-----------\n\nNO I WILL NOT HLEP YOU. DO YOUR OWN WORK! your asking someone to do your work for you.