Thrown off by this physics question...?
818855
You are correct that if the bell and passenger are both on the train then the freq would be the same.  But, I think the intent of this question is to say that the bell is on the station.  It says 'train station bell'.  Since the bell is on the station and the person is on the train, there is a relative motion between them.
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
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
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
85739
somebody help me with this! i think i just lost point for trying to understand the question.
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
941720
Doesn't that have something to do with Einstein's e=mc^2?  It sure looks like it.  The rest of seems to be an example of an explanation of the inefficiency of practical mass to energy conversions.  I could be wrong.\n\nOh, by the way:  what are you asking?
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.
711513
Hi!  You are right.\nIf they have equal masses and are elastic then  b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :\nVf= final speed;   Vcm=center of mass speed ;  Vi=initial speed.\nThus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=\n=m1*Vi/(m1+m2)  => m1=m2.    \n  " a)" isn't possible because the total momentum must be preserved.\n          Hope it helps!
641452
opposite force being applied
922486
Is this a trick question?\n\nLOL.
938772
Of cause, help.\nthe force of gravity is mM/R^2 = 550 N\nThis is for R = 3400km\n\nIn case we take the object up for 12 km, the distance from the centre of the planet will be 3412 N.\n\nand its weight will be mM/(R+12km)^2 = Wn\nWe can find  Wn   just from these two equations:\nfrom the first equation we shall express \nmM = 550(N)xR^2\nthan  Wn = 550 (N) x R^2/(R+12)^2 = 550(N)x3400^2/3412^2 = \n550 (N) x 0.996483^2 = 550(N)x0.9929783 = 546.138 N\nThat is the answer
380004
I don't understand.  Can you explain the problem a little better?
97560
Are you referring to a fluid mechanics problem, or?\n\nCan you restate the question?