When bullet manufacturers rate a bullet's energy as 400 ft-lbs is that refering to the total work done or is that just the instaneous torque? I'm a little confused I know a handgun doesn't recoil with 400 ft-lbs of force otherwise it would fly out of the shooter's hand or knock the shooter down.
213030
You must not think about forces but impulses p=mv! The sum of all impulses must always be the same (0) and therefore m1v1=m2v2. If you observe the velocity of the gun you get v1=m2v2/m1. Since the mass of the gun is so much bigger than the mass of the bullet you get a smaller velocity of the gun (and lets not forget the man holding the gun and is adding to the impulse)!
11996
reverie's wrong.  Most mass actually comes from kinetic energy, not rest mass.  you can correcly give energy (including heat energy) the same units as mass so it makes sense to say "one pound of heat"... but force is something different entirely.
43281
When masses collide, they exchange energy and momentum.  I'm not sure what you mean specifically by "power division", but you might be referring to this exchange of energy, since power is defined as energy transferred per unit time.  If so, just know that the important thing is that the total amount of energy and momentum is conserved, that is, remains constant.\n\nAddendum-  What the next correspondent says about elastic and inelastic collisions is correct; it makes a difference in how energy is transferred.  In every collision, total energy is conserved, as I said, but in elastic collisions, mechanical energy is conserved as well.  That is, none of the energy is converted to other forms of energy like heat, light or potential energy of deformation.  In inelastic collisions, momentum and total energy are conserved, but some of the mechanical energy is converted to those other forms.
382642
The bullet rises until all its kinetic energy is converted to potential energy. \n     Its kinetic energy at start is (1/2)mv^2 = (1/2)*300*(747*5/18)^2. Here the 5/18 term is the conversion factor from kmph to metres per second.\n    When it reaches the top most point 'h', its potential energy will be mgh\n\nThus (1/2)mv^2 = mgh\n\nTherefore, h=v^2/(2g) = (747*5/18)^2 / 2*9.81 = 2194metres = 2.194kms
222354
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
397225
Bound's hubby here:\n\nA grain is a very precise unit of measurement. a grain is 1/7000th of a pound ... 7000 grains to a pound.\n\nIn rifle and pistol ammunition, the bullet weight is expressed in grains, as is the powder charge. For example, a typical .308 Winchester target round would be a 168 grain Sierra MatchKing loaded on top of 41.5 grains of Accurate Arms 2520 powder.
637171
My friend, in my opinion, you have one of the finest .44 magnum revolvers ever made. The 329PD was designed to be the strongest and lightest weight .44 mag. revolver on the planet. With a titanium cylinder and the scandium alloy frame, which by the way has a strength to weight advantage of 70% over steel, it is safe to shoot any weight or  type of round you want. But like any gun, if you give it a continuous diet of hot ammo, it will loosen up sooner or later.\nShooting a .44 mag. 320gr.HC bullet @ 1175fps, you must be doing some serious handgun hunting!\nSo, go ahead and fire away. Have fun and be safe!!!
496395
The bullet's momentum before is m*v.\n\nThe spinning rod's momentum is I*w. Where w is in rad/s. But what is I? The rotational inertia of the rod, with the projectile stuck on the end, is I.\n\nThe rotational inertia of the rod is \nIrod = M*l^2/12 \nwhere l = length (in this font, capital I (i) looks identical to small l (L) - at least on my computer). You can consider the projectile to be a particle of mass m, so you can add the rotational inertia of a partical a distance l/2 from the pivot to Irod. So\nI = Irod + m*(l/2)^2\n\nSo the spinning rod's momentum is (Irod + m*(l/2)^2)*w. From w, I trust you can get the linear speed of the end of the rod.
381081
The force applied to the beast (by the bullet) would have to be calculated by the angle/arc of it's descent, this new force differential may cause a new vector.  ie  If it was falling straight down and you fired straight up...no change.....if it were coming in at 45^ angle down, the bullets force will slow it slightly, and it's arc down will have increased due to the continuing force of gravity...with less momentum.  That is...also only if the head stays attached....if not then...the two bodies must be calculaated by the individual resistance/force of the bullet....    In the end...take one step to the right...after your shot...and watch out for falling bits and pieces.
580515
This is a very good question.  Basically, the reason that we say kW and do not say kW/hr, is that calories and British Thermal Units (BTUs) are units of ENERGY, while the Watt is a unit of POWER.  The unit of time is already incorporated in the Watt.\n\nPOWER is defined as the rate at which WORK is done.  WORK is, in general, a transfer of ENERGY.  So POWER is the rate at which ENERGY is transfered from one object (or system) to another.\n\nThe capacity of a burner is stated in terms of POWER, that is, the rate at which it can transfer energy from the fuel (chemical potential energy) to the objects being heated.\n\nA Watt is defined as 1 Joule / second, (the Joule (J) is the basic SI unit of ENERGY).  1 Watt (again, a unit of POWER) is therefore equivalent to 3600 J/hr, since there are 60*60 = 3600 seconds in one hour.\n\n-------------------------\n\nFor your example, the power rating of the burner is 30kW.\n\nPower = 30kW = (30,000 W)*[(3600 J/hr)/(1 W)] = 108,000,000 J/hr = 108MJ/hr\n\nWe now use two conversion factors:\n\n1 kilocalorie = 4184 joules\n1 BTU = 1055.056 joules\n\nUsing these equivalencies, we can calculate the power in kcal/hr or BTU/hr:\n\n(108,000,000 J/hr)*[(1 kcal)/(4184 J)] = 25,812.6 kcal/hr\n       (this is pretty close to 25,800 kcal/hr stated on the burner)\n\n(108,000,000 J/hr)*[(1 BTU)/(1055.05585 J)] = 102,364.2 BTU/hr\n\n-------------------------\n\nYou stated that you are using Natural Gas with calorific value 40MJ/m^3.  (I found 38MJ/m^3 on Wikipedia, but this rounds to 40, so I'll go with your number for the example.)\n\nWe have:\n40MJ/m^3 = 9560.2 kcal/m^3  (using the above conversion factor)\n\nSo the consumption of Natural Gas is\n\n(25,812.6 kcal/hr)/(9560.2 kcal/m^3) = 2.7 m^3/hr\n\n(This rounds up to 3 m^3/hr, so your answer above is correct.)\n\nPlease note that 40MJ/m^3 is NOT equal to 11 kW/m^3, because the former is an ENERGY density, and the latter is a POWER density.\n\nHowever, 40MJ/m^3  IS approximately equal to 11 kWH/m^3, which I assume is what you meant to write.\n\nUsing this method gives the same answer:\n\n(30 kW) / (11 kWH/m^3) = 2.7 m^3/hr  (approx. 3 m^3/hr)\n\nAlso, as a matter of notation, the term "LPG" (Liquid Petroleum Gas) is typically used for either Propane or Butane in liquid form, NOT for Natural Gas.\n\nI hope this explanation is helpful for you.
294768
There are many different measures for destructive ability.\n\nTypically this is measured in units of energy.  For bullets, they measure the kinetic energy of the bullet which is then given in Joules (a measure of energy).  For atomic bombs the common measure is kilo-tons which is a measure of energy as well that can be converted to Joules (just a whole lot of them).  A kilo-ton is about 4,200,000,000,000 Joules.  A .45 bullet has about 450 Joules.\n\nLasers also impart energy on their targets.  The measure is typically in megawatts.  A megawatt is 1,000,000 Joules per second that it is applied.\n\nThe number of Joules doesn't tell you quite enough though.  The reason is because a single Joule of energy well aimed, could kill a person, but if you go out running in the sun, you will receive thousands of Joules in a short period of time from the Sun (which is obviously not destroying you).\n\nThe amount of energy that it takes to heat a cup of coffee is considerably more than the amount of energy it takes to fire most bullets, yet you can safely drink the coffee and the energy contained (as heat) within it.
11451
nothing's wrong with the above and your conversion is correct.\n\ndoesn't seem like very much eh?  the reason is b/c of what i explained in the other question.. if you plot a graph of the force versus time, it will look like this:\n\n_____|___|___|___|___|___|___|___|_____\n\nthose sharp vertical lines represent the bullet impacts, and the rest of the time the force is zero.  if you average over this time span, the number turns out to be very low.  but the height of those peaks is very high because so much momentum is exchanged in a very short time!  so force is kind of misleading.  in physics the more meaningful way to talk about collisions is to talk about the momentum exchanged in a collision, which is called an "impulse" because it happens over such a short time.\n\nput another way, the bullet has a momentum of m * v, or .003 kg * 740 m/s = 2.22 kg*m/s.  think about what that means.  it's like getting whacked in the chest with a 2.22 kilogram (around 4.5 lbs) object moving at 1 meter per second - pretty heavy and pretty fast!  \n\nand that's just one bullet.  think about 8 of them hitting you every second.  furthermore, 3 grams seems like pretty low estimate for the mass of a bullet.   that's something like a .22 or a .25.  consider that a .44 magnum bullet, which is the sort of gun that actually can knock you off your feet, has a mass of 250 "grains" which is closer to 16 grams.  that's more like getting hit with a 10-kg object.  that's like someone winding up and throwing a 22-lb weight at you!\n\nso there you go.  the averaging and the short time interval for the impulse is what's misleading you.
578143
Friction aside, think of it like this:\nwhen the bullet is fired it has a certain amount of speed.  every second that speed is being reduced by gravity by 9.8 m/s.  So when it reaches as far up as it will go it stops because gravity has subtracted 9.8 from from the original speed every second until the speed is 0.  Like taking out money from an account.  So now you can spend all the money you have-it's got potential.  Then it starts moving toward earth.  gravity is still working at the same rate. so it picks up speed at the same rate that it lost it last time. And it's the same distance, so it'll take the same amount of time.  So now you're depositing the money into a different account.   You only have the money you took out of the first account, so that's how much the second account will have.\nI hope that explains it for you.
492206
Recoil, in common everyday language, is considered the backward kick or force produced by a gun when it is fired. In more precise scientific terms, this force is equal to the derivative of the backward momentum resulting when a gun is fired.\n\nThe backward momentum is equal to the mass of the gun times its reverse velocity. This backward momentum is equal to the sums of the two forward momentums by the law of conservation of momentum and is due to and equal to the projectile's mass times its velocity added to the mass of the escaping gases, burnt propellant, unburnt propellant, and other gases, times their respective velocities. \n\nFor your case, we don't know enough information about the gases and other by-products of the firing. However, if we assume they are negligable, we can focus on the bullets.\n\nConservation of momentum:\nMh * vh1 = Mh * vh2 + Mb * vb\n\nwhere;\nMh is the mass of the helicopter\nvh1 is the initial velocity of the helicopter\nvh2 is the velocity of the helicopter after the firing\nMb is the mass of the bullets\nvb is the velocity of the bullets\n\nDepending on the accuracy of the answer you want, you may want to consider that the mass of the helicopter will change as the bullets leave the gun.\nMh * vh1 = (Mh - Mb) * vh2 + Mb * vb\nhowever, you don't have enough information to solve it in this case. You would need to know the velocity of the helicopter after the firing (vh2). Don't worry about it, I am probably just confusing you.\n\nBack to the question:\nMh * vh1 = Mh * vh2 + Mb * vb\n\nMh * vh1 - Mh * vh2 = Mb * vb\n\nMh * (vh1 -  vh2) = Mb * vb\n\n(vh1 -  vh2) = Mb * vb / Mh
434566
Work, or energy, is the product of force and distance.\n\ne = f d\n\nIn this problem, the distance doesn't change, so the work isn't dependent on the distance at all. But the weight does change, and since weight is the force (f) here, the amount of work changes in proportion to the weight.
237582
OK, the barrel´s mass is 40Kg (notice: its mass is 40 Kg, not its weight). \n\na) So, your formula is Weight = mass(m)*gravity´s acceleration(g)= 40Kg*10m/s^2 = 400 KG*m/s^2 (=400 Newton)\n\nb) Work = force * distance\nSince work is calculated only for the vertical, and the force equals the weight, you have:\nWork = 400N*5m = 2000 Joule\n\nc) Eff = 250N*5m/2000J*100 = 0.625*100 = 62.5%\n(You got 12.5% because you forgot to multiply 250N for 5 meters. Remember that efficiency is a relation of works, and work is force * DISTANCE).\n;o)
80639
Conservation of momentum.  Newton tells us that for every action, there is an equal and opposite reaction, such that in a collision, the net momentum remains the same.  When you realese gas backwards, the momentum imparted on the gas away from the space craft needs to be ofset by imparting momentum upon the astronaut in the direction of the space craft.\n\nWhen you fire a gun, the momentum of the bullet forward is equal to the momentum of he gun backward.  Since the gun is much bigger than the bullet, it doen't move as fast, and the force is spread over a larger asrea (your hands and shoulder).  But the kick backwards is the same sort of principle: the equal and opposite reaction.