Physics Multiple choice question?
829988
A) isn't likely to be correct unless the person was deafened by the sound. For a person with undamaged hearing there is no "too large" an amplitude for hearing the sound. The sound could be too loud to make any sense of it, but that's another issue.\nB) is plausible. Most humans hear up to about 20 kHz when young, and this cutoff frequency becomes lower with age and/or cochlear damage.\nC) is the most interesting possibility. The sound would be inaudible if only transverse waves reached a sensing point of zero size, but such a point couldn't have a directional characteristic so it would also receive longitudinal (compression) waves. A finite-sized sensing area consisting of a flat flexible membrane with one-sided exposure to the air (like the eardrum) could be positioned parallel to the direction of propagation and compression. Then it would be swept by transverse waves, and pressure differences would travel across its outside surface and be transmitted into the inside space (middle ear) and heard. (Note this would not be true for a flat two-sided sensing area, which would have canceling pressure differences on the other side.) In addition, the external ear directs the sound toward the eardrum so longitudinal waves reach it. So C is ruled out.
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.
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.
14955
(a) it's an inelastic collision.  An inelastic collision is when they stick together, and an elastic collision is when they bounce off of each other\n(b) p=mv  momentum= mass x velocity.  p= 30 m/s x 500kg = 15,000 kg*m/s\n(c) well, momentum is conserved, so the we would use the formula from before and the momentum(p) from before, except that now the mass is different because it is the combined masses of the two vehicles (300+500=800) , so you can solve for the new velocity.\np=mv  15000 = 800 x v\n\ndivide both sides by 800 to get  v=15000/800 = 18.57 m/s
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
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.
816137
a) v^2-u^2=2as\n    v^2-52.4^2=2*(-9.8)*28.5\nv^2= 2745.8-558.6\n     =2187.2\nv=46.76m/s\n\nKinetic Energy=1/2mv^2\n                     =1/2*46.9*46.76^2\n                      =51,290 Joules\n\nb)height=28.5-5.06=23.44m\n\nv^2-u^2=2as\n\nv^2-52.4^2=2*(-9.8)*23.44\nv^2=2745.8-459.42\n     =2286.4\nv=47.8m/s
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
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
475727
the answer to the first question is 5 m/s.\nF=ma; a=F/m=200N/40kg=5m/s\nthe second answer is 1m/s.\nF=ma; a=F/m=2N/2kg=1m/s\nthe third answer is 1:4.\nF=ma; m=F/a\n(1) M=f/a=F/a=m\n(2) M=f/a=4F/a=4m
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
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
453032
three answers above are not reliable.i strongly suggest you to ask your teacher.
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track