physics help please!!!?
820184
I think you should break things down and think of the question like this.\n\nThe heat for a process is 250.3J, Its doubled with the same temperrature difference, the temperature difference is 5C.
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
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
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
660156
http://www.physlink.com/education/askexperts/ae25.cfm
703117
here u go!\n a):\nuse the principal of F=ma to find the frictional force.  \nIn order to do this, you need to determine acceleration, so use the kinematic formula: v^2=vo^2+2a(x-xo).  \nwith this equation, you should find a to be equal to -1.5m/s^2.  \nPlug this answer into the F=ma equation, with m equal to .090kg.  Since Friction is the only force in the horizontal direction, F is equal to the frictional force.\nSolving for F=(.090)(-1.5)\nF=-.135N, so its magnitude is .135N\n\nb.):\nOne physics principal states that a frictional force is equal to mew (the coefficient of kinetic friction) times the normal force.\nSince you now know the frictional froce, just solve that equation.\nThe normal force in this problem is equal to the opposite of the gravitational force, which is (9.8)(.090)=.882\nSo, when you have the equation \n(.882)(coefficient of friction)=.135\njust divide .135 by .882 and you will get the coffecient of friction to be .153!!!!!!! yay!!!
284176
ok well his initial velocity is 0 and the distance is 402m and time is 4.913. you have to find the acceleration. well the only way i can think of doing this  without using these basic equations that are derived from newtons law (i forgot the derivation) is this.     you find the average velocity needed to go 402m in 4.913s. which is just 402m/4.913s = 81.823....\nthis is just the average velocity and because acceleration is constant, this means you can just find the average by doing  Vi + Vf / 2. you have Vi =0    so Vf/2 = 81.823........ so Vf = 163.6474.........m/s\nacceleration is  change in velocity /time. you have change in velocity  (Vf-Vi)   so divide it by time.   \n163.6474-0/4.913   =  33.3 m/s^2\n\n2. so the acceleration due to gravity is 9.8m/s^2. you know the distance is 10m and initial velocity is 0. so Vf = a x t\nso Vf = 9.8t. and Vf+Vi/2 = d/t    same things basically. so Vf/2 = 10/t  which means  Vf x t = 20\nVf = 9.8t\nVf x t = 20\nso then that means that (9.8t) x t = 20\n 9.8t^2 = 20\nt^2 = 2.04....\nt = 1.4285.....  so it takes her that long\nand now you just find Vf by putting this time in the equation  Vf = 9.8t = 14m/s. thats final velocity\n\nb. ok so this question confused me at first but it snot hard. once she hits the water (with speed of 14m/s), she travels for 2m before coming to a stop. so distance is 2m. Vf = 0. Vi = 14m/s and you have to find acceleration. \nso a = (Vf-Vi) x t = -14t\nso you need time now. you do average velocity  again\nVi + Vf / 2 = d/t\n14/2 = 2/t\n7 = 2/t\nt = 0.2857.... s\nthen put it in the other equation and you get a = -14t\nso a = -4m/s^2.\nforce = mass x acceleration\nmass is 61kg so you do 61kg x -4 = -244 N\nthe water exerted 244N against the movement (thts why its negative).\n\nso hope this helped. let me know if you need more help or whatever. laterz
269311
assuming that force is applied in the same direction that displacement....\nwork = force * displacement * cos(angle between force and displacement)\nwork = (10N)*(5.0m)*cos(0)\nwork = 50 Joule
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
1093836
Centrifugal force and gravity.
170255
It is much easier to just give you a link instead of explaining it:\n\nhttp://www.shef.ac.uk/physics/teaching/phy107/solvegray.html
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
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
883480
this seems an easy question but most of the student get it wrong since it is so easy to mess up your mind. it was a part of an exercise for my University exams. it makes you realise what is going on with spring ...just consider it as a opposing force. the body will drop and will expirience g --> it will accelerate. at the same time it will also expirience the sping;s opposing force.\n\nit will accelerate till the force become equal. this is the importnat factor. weight will keep on accelerating the body till it is completely cancelled by the spring force.\n\nso when \nFs = Weight then Velocity is maximum. \nso kx =mg =>\n\nx = mg / k.  (at this displacement the velocity  will be maximum)\n\nat this specigic point due to the theorem of conservation of energy we get that.\n\nkinetic energy + sping stored energy = potential energy at the start.\n\nso \n1/2mu^2 = (m*g)^2 / k => (be carefull mass = 76.7/1000kg)\nU = \n\nU = sqrt (( m*g^2)/k)) => u = 0.361m/s 9\n\nIF THE ALGBRA IS CORRECT\n\nNOW AFTER THAT POINT the body will still travel due to inertia and the springs force  larger now than gravity wll slow it down and bring it at rest \nat that point all the potential energy (due to height difference) will be transfornmed to springs stored energy therefore,\n\n1/2 k xmax^2 = mgXmax =>\n\nXmax = 2*m*g / k =>\n\nXmax = 0.02658m \n\n\n( i am at work and i believe the algebra is correct. the main point is to realise where certain things happen.)
772638
try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses