a coin is placed on a vinyl stereo record making 33.33 revolutions per minute.?
965799
Dude. What textbook from the 70's are you working out of? A 33 record? I think my grandma has some of those!\n\nNot to continue on the outdated of the details of this question, but ask someone 40 or older if 15cm from the center of a 33 would even still be on the vinyl! I don't remember if it was 33's or 45's that were the little ones. I'm thinking 33's were the larger- so 15cm would put the coin right the outer edge.\n\nIt would make a difference, so I'm not just being smart a**.
636056
one coin is not a nickel. the other one is. the answer is a quarter and a nickel
1142512
rotational sensor.. which is basically an inductive prox with a timming circuit to calculate revolutions.
293352
Your question is flawed...\n\nThe drive spins the disc faster while reading on the outer area, and slows down as it gets closer to the inner area.\n\nIt's not like an old record player where there is one constant speed.
811425
This is more of a geometry/arithmetic problem than physics.\n\nLet's reason through it.\n\nThe reel has a radius of 3 cm, so its circumference is 2 x pi x 3 cm.\n\nThe line is reeled in at a rate of (2.5 / 9.5) m/s, \nor 100 x (2.5 / 9.5) cm/s.\n\nBy doing these calculations, you will have the centimeters per second at which the line is being reeled in, and the centimeters per revolution at which it is being reeled in. If you divide centimeters per second by centimeters per revolution, the result will be the revolutions per second at which the reel is turning. This is the angular speed that you need to calculate.
201872
The normal commercial disc is engraved with two sound bearing concentric spiral grooves, one on each side of the disc, running from the outside edge towards the centre. Since the late 1910s, both sides of the record have been used to carry the grooves.\n\nThere is an area around 6 mm (0.25″) wide at the outer edge of the disk, called the lead-in where the groove is widely spaced and silent. This section allows the stylus to be dropped at the start of the record groove, eliminating the risk of damage to the recorded section of the groove when the stylus head is dropped carelessly onto the record.\n\nBetween each track on the recorded section of a LP record, there is usually a short gap of around 1 mm (0.04") where the groove is widely spaced. This space is clearly visible, making it easy to find a particular track.\n\nTowards the label centre, at the end of the groove, there is another wide-pitched section known as the lead-out. At the very end of this section, the groove joins itself to form a complete circle; when the stylus reaches this point, it circles repeatedly until lifted from the record. Automatic turntables rely on the position of the arm, as it reaches these more widely spaced grooves, to trigger a mechanism that raises the arm and moves it out of the way of the record.
20548
The cheapest way would be putting a mic up to your stereo with the record playing, and with some kind of audio converter (M-AUIDO makes nice ones) hook it up to your computer.
862904
28 beats per minute\nThe site below is cool, it shows a diagram as an example...
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
828009
One revolution = 2pi radians.  One minute = 60 seconds.  So 1 rev/min = 2pi/60 rad/s.  Multiply 11.2 rev/min by 2pi/60 to get 1.17286 rad/s.
609168
Excellent question!\n\nBecause film is composed of 25 frames per second the much-faster revolution of the holes in the wheel rim gets out of phase.
894975
Per minute - if it's a mobile.
730990
You would have to take it to a coin dealer to find out. They can grade its condition and look it up. Chances are, it isn't worth more than 25 cents, but it is worth at least a penny!
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
1128414
It looks like you have a 1932 Five Lati Crown slver coin.\n\nBelow is a link to a coin dealer that was offering the same coin for between $23 and $42 US.  I don't jknow if those are current prices.\n\nWith all coins - the value of the coin is related to the condition / wear of the coin.
1124301
Are you trying to figure out when it stops? \n\nIf so, look at your units:\n\nYou have meters per second squared, and meters per second... To get to seconds you're going to invert the meters per second squared and then multiply... \n\nYour answer is 7 seconds.
986802
91-92 MPH on average. I think he's been clocked at 94 before. And no, it can't go back in time. Without a time machine, anyways.