Physics PLS HELP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
965199
Firing the weapon vertically...\n\nV^2 = Vo^2 + 2gh\n0 = 600^2 + 2(-9.81)h\nh = 18.348 km ; where h is the maximum height reached by the bullet\n\nThe work done by firing this bullet to reach the max height is...\n\nW = mgh\nW = (5 gf) (9.81 m/s^2) (18.348 km)\n\nBTW, 5 gf? What the fcuk is that? I'll just assume this as 5 grams.\n\nW = 900 N-m or Joules\n\nPower is the rate of doing work. Power therefore is equal to...\n\nP = W x bullets fired per min.\nP = 900 x 250 bullets / 60 secs\nPower = 3,750 Joules per sec or watts.
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
707702
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)
227728
Very simple. start out asking yourself what information do i know about the problem. Anything from masses and forces to symmetries of the problem. Next Ask yourself what additional information you need to solve the problem, values, formulas, etc. from there it should all fall out with ease.
53841
The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
539746
F = ma\n\nF - Force\nm - mass\na - acceleration
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.)
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
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.
1093339
No one on here is going to actually DO your homework for you.. but here is a place for you the get help -\n\nhttp://www.themathpage.com/aPreCalc/logarithms.htm\n\nhttp://en.wikipedia.org/wiki/Logarithm\n\nhttp://physics.mtsu.edu/~wmr/log_2.htm
754118
E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
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
1152255
www.physicsclassroom.com
877424
In the vertical direction earth is pulling the object downward with a force.\n\nAnd all objects are pulled down by the earth with an acceleration given by g = 9.8 m/ s^2.\n\nFirst we consider the vertical motion alone.\n\nIn the vertical direction \nthe initial velocity u = 0.\nthe acceleration  a = g. \nThe time t = t to be found.\nDisplacement in the vertical direction   s  =   180 m.\n\nWe use the equation s = ut + 05 a tt.\n\nUsing this we find t = 6 second.\n\nNow let us consider the horizontal motion.\n\nThe object is released from the plane which was moving with a speed of 20m/s.\n\nThere fore all objects which was in the plane will also have a speed of 20m/s in the horizontal direction.\n\nHence in the horizontal direction the initial speed is 20m/s.\n\nThe acceleration in this direction is zero since no force is acting in this direction.\n\nThe time t is the same as 6 second.\n\nDistance we have to find.\n\nWe use the equation s = ut + 05 a tt.\n\nSince a = 0 the equation reduces to s = ut. = 20 x 6 = 120m\n\nThe horizontal displacement is 120 m.
879813
you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
870625
each of the particles would have 1/2 the energy of the spring, 30 j each.  They would move at different speeds, c moves slower than d, but energy 1/2 mv^2 would be the same.