help me, physics word problems suck?
819223
If I understand correctly, one revolution of the disk is equal to one period.  You need to know the time (T) taken for one period.\n\nFrequency (f):\n f = 1/ T\n T = 1/ f\n T = 1/ 2200\n T = 0∙000 454 545....\n T ≈ 454∙454 µs\n\nAngular speed (w):\n w = Θ/ t\n w = 2π/ 454∙454 x 10^-6\n w = 6∙283 185 307.../ 454∙454 x 10^-6\n w = 13823∙007 68...\n w ≈ 13,823 rad/s\n\nIf the period is 20 for one cycle then:\n w ≈ 13,823 / 20 rad/s\n w ≈ 691∙15 rad/s
943651
I am pretty sure that if you looked in your textbook, it would give examples of this kind of thing, and from that you should be able to figure it out on your own. You're a smart kid, you are taking physics instead of some easy A class, now act like it, stop asking others to do your work, and try to solve it on your own.\n\n\nLong Live Jambi
610467
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
676519
When I did a Physics problem wrong.True.
951948
When I do word problems, I break it down into manageable chunks.  After reading the entire problem once to get a sense of what I'm dealing with, I re-read each sentence one at a time and try to write down in mathematical terms what each sentence is saying.  Often each sentence is only a portion of the overall equation, but I find if I tackle it like this, I can piece together the overall equation by looking at all the partial equations I got from each sentence.\n\nHope this helps a little.  Good luck!\n\n~ ♥ ~
383748
Facts to remeber while solving the physics problem\n1. A force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1 Newton.\nWe will also be using the principles based on Newton's Laws of motion.\nEarth attracts a mass of one kilogram with a force of about 10 newtons.\nIn Physics weight and mass are different. The mass of a body does not change, the weight will depending on the gravitational pull acting on it.\n   \nF = m x a       \nWhere    m = mass (kg) \n      F = force (N) \n      a = acceleration (m/s/s) \nTo calculate the downward pressure on the bottom of the bucket .\nWe first calculate the force on a 1 Kg bucket (given) due to 30 cm(change to metres 1 m = 100cms) of water\nF =  1 x 0.3 x 0.3 = 0.09N\nTherefore total downward force on the bottom of the bucket is the addition of the two.\nTotal downward force = 10N + 0.09 N = 10.09N\n\nTo calculate weight of full bucket.\nWe know bucket weighs 1 Kg\n12 litres of water = 12 kg\ntherefore total mass = 13 kg\n1 Kg = 10N\nTherefore 13 Kg = 130N\nWight of full bucket = 130 N
1014316
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
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...
69843
work = force times distance\nW=Fs
782037
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
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.
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
773883
Stop for a minute... Look over your question. Note 2 things\na. 77.2 AU\nb. An icy object\n\nNow start over slowly and find your mistake...YOU CAN DO THIS!!!