physics question need help!!!! + need the work or explination for it?
953154
Acceleration\n    a = w^2 r\n    w = 2pi / T\n    w = 2 x 3.14 / (365.25x24x60x60)\n    w = 1.99 x10^ - 7\n    a = (1.99 x10^ - 7) ^2 x 1.5 x 10 ^11\n    a = 5.95 x 10 ^ -3 m/s^2\n\nForce\n\n    F = mw^2r\n    F = 6.02 x 10^24 x (1.99 x10^ - 7) ^2 x 1.5x10^11\n    F = 3.58 x 10^ - 22 N\n\nSource of this force:\n    It is the gravitational force of attraction between the sun and the\n    Earth. It is given by Newton's law of gravitation.
735664
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
969486
I just helped someone else with their physics question, so I'm alittle tired of it.  But, I suggest you make an equation sheet...physics is all about equations.  So, if you have a piece of paper with your equations, all you have to do is look at the sheet and plug things in...
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
742117
work = f*d\nso we just need the force since we know d=5\n\nthe vertical force on the surface is:\nmg-sin(15)*250\n=90*9.8-sin(15)*250\nto compute the frictional force, multiply by the coefficient of friction\nand by the distance for the work\nF=5*.2*(90*9.8-sin(15)*250)\n=817.3\n\n\n\nj
920147
do your own homework. at least try to make the question not sound like a homework question
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
735831
Did you put your entire homework assignment on here? I noticed you posted a bunch of science questions just now.\nIf you don't have your textbook handy [you really should be looking up these answers so that you can remember and understand them when you have your test/exam], go to the links below and look them up. The 1st is a down loadable physics textbook. The 2nd has animations of physics.
265191
Work = Force x Distance\n\nIn this case, since the shot is going up, i.e. opposite the force, the work done on the shot is negative.  Another way to look at it, if up is our positive direction, the distance will be positive but the force, i.e. the weight, is negative.  Note that the distance over which the weight acts is the difference in height, not the entire height of the shot relative to the ground.\n\nW = Force x distance = Weight x distance = - 71.6 N (2.09 m- 1.60m)\n\nI'll leave the arithmetic to you.\n\nTo answer the second part, just notice that the change in potential energy of the shot will be equal in amount to the work done on it.  Just take away the negative sign, since its potential energy is increasing. Another way of looking at it is to say that the work done on the shot takes away its kinetic energy, hence the negative sign, and converts it to potential energy.\n\nSince you're only concerned with the change in potential energy, you only have to concern yourself with the difference in height.  If you want to use the ground as the reference level, and calculate potential energy, you get\n\nDEL PE = PFf - PEi = Weight x final height - weight x initial height\n\nDel PE = 71.6 N (2.09 m- 1.60m)\n\nSo it works out to the same thing.  Yippee.
695472
Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
984947
W=F*L – work done by F=145N on distance L=7.2m,\nU=k*m*g*cos(a)*L, work wasted on friction, where k*m*g*cos(a) is force of friction, while\nk=0.31, a=11.7deg, g=9.8m/s^2, m=10.3kg.\nY=m*g*L*sin(a) work done to hoist m to height of L*sin(a) meters.\nE0=m*v0^2/2 initial kinetic energy of m, v0=1.3m/s \nNow energy balance: E0+W=U+Y+E1, E1 is final kinetic energy of m E1=m*V^2/2, V is final speed of m. Thus kinetic energy change dE=E1-E0=W-U-Y.\nAs E1=E0+W-U-Y=m*V^2/2, then V=sqrt[2*(E0+W-U-Y)/m], \nI hope you cope with numeric calculations, all are SI units
950937
You're in luck.  I teach physics and I know exactly how to answer this question (it's based on the work-energy principle).  It's pretty simple, really.  But it would be best if we discussed it by messenger.  Yahoo Messenger me, my ID is\nfortitudinousskeptic\n\n- Kevin
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m
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
696021
OK, you already know (a), which is the total energy. At the peak displacement (which is the amplitude), all the energy is potential energy PE, in the form of spring deformation. PE = .5*k*x^2 = .5*35*.04^2 = .028 j. At other times, the energy is divided between PE and kinetic energy KE, and PE+KE=.028. So for (b), KE = .028-.5*35*.01^2 = .02625 j. Then the velocity = sqrt(2*KE/m) = 1.025 m/s. For (c), KE = .028-.5*35*.03^2, and for (d) PE = 5*35*.03^2.