I need professional help on this physics question.?
961109
It's good to see people actually thinking about what the terms they're using mean, rather than just plugging them into equations.\n\nBasically, the place where you seem to be confused is here:\n\n"In gas such as water vapour, intermolecular forces of attraction is weak and hence it has low PE"\n\nActually, this means the exact opposite, that the system has very high potential energy. Think about gravity for a minute - if a rock placed at, say, 1 foot above the ground were to fall to Earth, it would not release very much energy. Conversely, if the rock were placed 1 mile above the Earth, it would release a considerable amount of energy. The higher the rock is, the greater it's potential energy. However, the further the rock is away from Earth, the weaker the force of gravity acting on it - thus weak gravitational attraction corresponds to higher potential energy. Similarly with the particles in a gas - if they were to "fall" into a partially bound state such as that of a liquid, they would release energy, and thus the unbound state where the particles are too far away from each other to form bonds is the higher energy state.\n\nThis is not to say that a weaker force always corresponds to a higher energy state - if the reason that the force is weaker is that there is less mass/charge/whatever producing it, that really will lower the potential energy. But in the transition from a liquid to a gas does not change the charge of any of the particles - the reduction in intermolecular forces comes from the fact that the particles are moving out of each other's electrical potential wells, a process which, just like lifting up a rock, raises the potential energy.
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
344596
Wow, that is a very good and tricky question. If you really want to know please send me an e-mail and I will tell you the most recent answer proposed by the science community.
124298
Answer to #2\nt=1.565 seg\nV= 2.556 m/seg\nt=1.565 seg, the same.
860846
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
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!!!
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
380004
I don't understand.  Can you explain the problem a little better?
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
93876
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
190569
Only if you show your work and do not expect them to do the question for you; only to provide hints and guides.\nhttp://www.physicsforums.com
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...
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s