Using the relationship Fc=v^2/Rp^2 determine the centripetal force of an object traveling at 1,777 m/s at a radius of 200 miles. Caution: do not mix units.\n\nPlease help me!?
136903
Given that 200 miles is 322 kilometers\nand assuming that m/s is meters per second:\n\nF=Mv^2/r\n\nF=M * 1777^2 /(322000)\nF=M * 9.8\n\nSo, the force will be 9.8 times the mass, whatever that is.
488802
F = m*r*w^2\n\nm = mass = 2kg\nr = radius = 0.5m\nw = angular velocity = 5/r = 10 rad/s\n\nF = 2/0.5/10^2 = 100 N
524607
[Edit -- two things.  #1 Oops, I made a mistake.  When I worked this, I used a mass of 4 kg instead of 0.4.  Now I'll go back through my answer and make the correction.\n\n[#2  There are two different methods here.  Look at both of them.  The <second> way is <much better> for this problem.  I think you'll see that.]\n\nSnow White has an answer, but she doesn't say how she got it.\n\nLet's start by getting the net force F = F1 + F2 = -0.6Ni + Nj with magnitude sqrt(1.36)N = 1.166 N and acceleration F/m = 2.915476 m/s^2.\n\nNow we have two ways to proceed.  This is linear motion with constant acceleration, and we know the direction ... tan theta = -1 / 0.6 = -5/3, so the angle theta in the second quadrant is 59 degrees north of west.\n\nAfter 1.6 seconds, the velocity is acceleration x time = 2.915476 x 1.6 = 4.66476 m/s in the direction indicated above.\n\nThe displacement s after 1.6 seconds is s = vt/2 or s = 1/2 at^2 = 4.66476 x 1.6 / 2 = 3.7318 m.\n\nTo get the position, we could use sines and cosines, but notice that tan theta above is -5/3, so the hypoteneuse of that triangle is sqrt(25+9) = sqrt(34).  Using that geometry, the position after 1.6 seconds is\n\nx = -3/sqrt(34) x 3.7318 = -1.92 m\n\ny = 5/sqrt(34) x 3.7318 = 3.2 m\n\nThat's your answer: (-1.92, 3.2) with velocity 4.66476 m/s and direction 59 degrees north of west.\n\n\nThere's another way to do it that may be better, especially when you're using vectors.  Do it by components, and keep the components separate.\n\nThe net force is F = F1 + F2 = -0.6Ni + Nj.\n\nThe constant acceleration a = F/m = -1.5i + 2.5j m/s^2\n\nThe velocity after 1.6 seconds is v = at = -2.4i + 4j m/s\n\nThe position after 1.6 seconds is s = vt/2 = -1.92i + 3.2j\n\nand this is the same answer we got by doing it the other way.  (You can, of course, get all the magnitudes and angles.)\n\nLooks to me that it's easier doing it by components.\n\nHope this is the same answer Snow White got.  This way, at least, you have two ways of doing it.
462603
F=ma, and \na=v^/r\n\nTherefore, F=mv^2/r\n                    =m*34^2/155\n                    =7.45m N\n\nThere are 3 forces acting on the string, the weight mg of the block in the downward direction, the centripetal force, F which we have just computed to be equal to 7.45m N, and the tension T in the string.  If you draw a diagram showing this 3 forces, you will find that the angle from the vertical made by the string is equal arctanF/mg, i.e. tantheta=7.45m/mg\n                                                 =7.45/9.8\n                                                 =0.760\n\nUse a scientific calculator to get the angle.\n\nThe centripetal force on the van will be proportional to the centripetal force on the block, and the angle theta will be the same as the angle of the banked curve if it were to be banked. Likewise the radius of the circle will be proportional to that of the  \nblock. The reason for this is that the velocity of the van is the same as the velocity of the block. Thus the terms in the formula F=mv^2/r for the van will be in direct proportion to those of the block, if we compare them with one another.
466450
You've missed a unit, but I'm going to assume the distance is 5.6 metres ...\n\n(Velocity) = (Work) = (Force) * (Distance)\n\n and\n\n(Force) = (mass) * (acceleration)\n\ntherefore \n\n(Force) = (mass) * (distance) / (velocity)\n\n... plugging in the numbers, we get   57.8 kg.m/s/s ... = 57.8 N.
41563
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
292500
I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
430532
I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
489517
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
531381
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
300228
V = (v + u) / (1 + uv/c²)\n\nTry that. The velocities are quite high so relativistic. Haven't tried it yet though.\n\nNevermind, the velocites aren't nearly high enough to be relativistic. If it was km/s it would work.
132613
well, using f=ma, and w=fd, we immediately get w=(ma)d\nW=work  (3650 joules)\nM=mass  257kg\na=accelleration =gravity =9.81\nd=distance (required)\n\nrearrange w=(ma)d\n\nd=w/(ma)\nDistance=work/(mass*acceleration)\n\n= 3650/(253*9.81)\n=1.47m
384970
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
29299
centripetal force is mv^2/r.\nso it comes around 1,777,777.778 N\n\nI need one more parameter that is coeff of friction.\nif that is available u can use\n\nv(max)= sqrt [(rg (u + tan o)/(1 - u * tan o) ]\n\nassuming frictionless\n\nV= sqrt (rg tan o)\nso \no = 66.2 degrees if g=9.8\nand 65.7 degrees if g= 10.
527202
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
521012
calculate the angular acceleration of the seed:\nw_i = 0\nw_f = 3.49 rad/sec\n\nalpha = (w_f-w_i)/t\nalpha = (3.49-0)/0.25\nalpha = 13.96 rad/sec^2\n\nconvert this to linear acceleration at the seed:\na = alpha*r\na = 13.96*0.060\na = 0.8376 m/sec^2\n\nThe maximum centripetal acceleration occurs when the record is moving at max speed so find the max centripetal acceleration:\na_c = (w_f^2)*r\na_c = (3.49^2)*0.060\na_c = 0.731\n\nthe 2 accelerations are perpendicular to each other so find the magnitude using Pythagorean:\nsqrt(0.731^2 + 0.8376^2) = 1.112 m/sec^2\n\nthen calculate the net forces:\nu*N = m*A\nu*(m*9.8) = m*1.112\nu = 1.112/9.8\nu = 0.113
345562
This is a very basic question and you need to get extra help if you are having trouble solving this in your physics class.\n\nAcceleration is force / mass.  (from F = ma).\n\nThus, for the astronaut\n\n120 = 79(a)\na = ~1.52m^2/s\n\nAnd the satellite\n\na = 0.182\n\nNow they travel in opposite directions (BOTH of them move). So calculate how far the astronaut travels, how far the satellite travels, and add them. Simple kinematics:\n\np = 2a(t^2) + v(t)