How do you calulate the probability of hit for a 25 mm cannon firing at a 2.3 m x 2.3 m target from 2000m?
438167
The probability will completely depend on the system.  If the system was perfect, and conditions were always favourable, then you would have P(H) = 1 all of the time.  Equations of probability are based on past performance to predict the future, and the more you know about the past, the better you can predict the future.  Predicting the probability of something occurring is really a matter of statistics, and I don't know how you could obtain an answer any other way given the parameters of the system.  If the system was perfectly designed, as I said you would always have a hit, and if it is not, then you cannot know what makes some bursts miss or you cannot control what you know the problem is, or you would fix it to get a perfect system.\n\nI would think that you are going to need historical data to statistically determine the probability.  Run a number of experiments, recording how many bursts were fired and how many has at least one of the 3 rounds hit the target.  Your "estimated" probability of a hit P(H), is then obviously number of hits divided by total number of busts fired.  Remember that conditions such as weather (wind) affect the system, so by controlling these variables you can construct several probabilities for varying conditions, or you can lump them all together by using a random sampling method.\n\nMore useful probably, a confidence interval can be stated to a certain degree of......well confidence.  If you wanted to get a range of probabilities of a hit that is 95% certain to occur for any given firing, then using a Gaussian Distibution table find 0.95, or whatever confidence level you want (0.95 is 1.96 btw).  then, since Gaussian is symmetric, you have:\n\nlet P = probability desired to get the interval for\nlet P_est = estimated probability from sample (the dividing thing from above)\nlet Sigma = variability\n\n-1.96 <  [(P_est) - (P)]/{(Sigma)*sqrt[{(P_est)*(1 - P_est)}/N]} < 1.96\n\nSorry thats not fun to write in a text box.  This is based on a mathematical model for probability.  There is a slight approximation in the above method, but it is often quite satisfactory, and the exact method requires an inequality quadratic solution.  Then re-arrage, remembering the rules of arithmetic with inequalities, and since you know everything except P bsaed on your data, you will have \n\n(some number) < P < (some other number)\n\nWhich is the range of probabilities for hits that, based on your data, will be valid for any given sample of firings you conduct, whatever percentage of the time that you selected as your level of confidence.\n\nThe more historical data (sample size) you have, the better you can predict the future probability, and the smaller the range of your confidence intervals will be.\n\nHope this is at least somewhat helpful to you.  This is just some of the basic statistics that I know.\n\nCheers,
813272
no its wrong\nthe correct answer is \n15 degrees and 75 degrees\nbecause the range of a projectile is same at \nangle A (say) and 90 degrees- angle A
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
851529
approximately 1107 feet i think\nof course this is neglecting air resistance and mass of the bullet along with loss of energy heating\n\nmy work:\nfind the time that bullet is in air:\nwe know that height is 5 feet and the initial velocity of the downward component is 0 because the gun was fired horizontally.\nusing second kinematics equation :\n intial velocity(time) + 1/2(acceleration)(time)^2 = distance\nwe fill in:\n0(t)+1/2(32.67ft/sec^2)(t^2) = 5 and solve for t which = .553283\nNow that we know the time bullet spends in the air we can find the distance by using time and the same equation but using all horizontal component.\nwe will in:\n2000(.553283)+1/2(0*)9.5532830^2 which equals =1106.57 or about 1107\n\n* = there is no acceleration since there is no gravitaional effect on horizontal component\n\ni made have made a mistake so please check
124804
For an uncertainty x of 0.01 (which your 99% accuracy would be) you need 1/x^2 samples which would be 10000.  This is true for a large population (>>10000).  I believe you still need that many for 20000 population but am only 99% sure.\nSource: Best recollection of a stat course 50 years ago.
886896
0.696850393"\n1"=25.4mm=2.54cm
877406
For the first one:\n\nz = (2725-2500)/150 = 1.5\n\nThe area would be 0.0668.\n\nFor the second one: You are given that X (the tensile strength) is greater than 2775, and you want to find the probability that X is greater than 2725.  This probability can be written this way.\n\nP(X > 2775 | X > 2725)\n\nRemember how to compute a conditional probability.\n\nP(A|B) = P(A and B)/P(B)\n\nso...\n\nP(X > 2775 | X > 2725) = P(X>2775 and X>2725)/P(X>2725)\n= P(X>2775)/P(X>2725) (The only way X>2775 and X>2725 is if X>2775.)\n\nConverting 2775 to standard units...\nz = (2775-2500)/150 = 1.83\n\nSo 0.0336/0.0668 = 0.503 approximately\n\nThe third one is zero.  Since a normal is continuous, the probability of any particular number is zero.  Think of trying to find the area of a line; it has length but no width.\n\nedit: jojo, you did not read very carefully.  This is not refering to a hypothesis test, and you did not write the second problem correctly.  It is a conditional probability.
923444
Short Answer: \n\nvert comp = .161 meters\nhori comp = .936 meters                        \n\nLong Answer:\n\nMb=mass of bullet\nMp= mass of pendulum\nv=velocity of bullet\ng=9.81 m/s/s\nh=vertical displacement\n\n\nBy conversation of momentum:\n\nmomentum of pendulum with bullet=momentum of flying bullet\n\nsqrt(2*g*h)(Mb+Mp)=Mb*v= (28/1000)*230=6.44 \n\nsolving that you can find h=0.161 meters\n\n\nBy the distance formula:\n\nsqrt(2.8^2+0^2)=sqrt( (2.8-0.161)^2 + (horizontal position)^2)=2.8\n\nsolving that you can find horizontal position = 0.936 m
976194
5,25rad/s = 2πf\nf= 0,836Hz\nVt= (2πr*f)\nVt= 2π*0,0600m*0,836Hz= 0,315m/s\nAc= (V^2)/r\nAc= (0,315m/s)^2/0,0600m= 1,66m/s^2\n\nBoth points have the same frecuency, since they have the same angular velocity. So, for point 2 you already have a velocity of 5,25rad/s.\nThen, Ac= (V^2)/r →  (4π^2*r^2*f^2)/r→ (2πf)^2*r\nTherefore, Ac α r.\nIf r is half, then Ac is half (1,66m/s^2)/2= 0,83m/s^2\nthe same for the velocity.\n Vt= (2πr*f), because the radius is half and Vt α r\nVt= 2π*(r/2)*f = π*r*f= π*0,0600m*0,836Hz= 0,158 m/s\nOr just (0,315m/s)/2= 0,158m/s
643054
Cannon had a range of several hundred yards. A 3 pounder ranged about 800 yards with solid shot, and 2 hundred yards with grape shot, maximum. At close range, loaded with shot, it could destroy an enemy company.
802288
the impuls (m*v) is conserved, so the impulse in one direction;\n0.150*20*sin(30deg)\nEquals the impulse in the other direction\n50*v_recoil\nthe recoil velocity is thus 0.03 m/s
794091
I'd suggest you first try the site listed below where you can find a downloadable spreadsheet that predicts velocity. You can deduce the author's equations from it. It's actually a time-stepping simulation rather than analytical equations. However, a bit of theory wouldn't hurt either. First, to keep it simple, let's assume the pressure is constant for T, the lesser of the time the valve is open and the time the projectile remains in the barrel. This is actually not a good assumption since many spudguns have pressure vessels that are smaller than or not much bigger than the barrel volume. Anyway, during T the spud is subjected to a force F=P*barrel area = P*pi*d^2/4. The net force accelerating the spud is F - friction force, so its acceleration A=(F-Ffriction)/M and its velocity v=A*T. I don't know the practice on valving. It may be that open valve time is always less than barrel transit time, but if not, you can compute the distance S that the spud would travel during the open valve interval T; S=0.5*A*T^2. If S>l, then T is the time spent in the barrel, which is SQRT(2*l/A), and you should recompute v. \nAnother assumption would be that each shot entirely empties the pressure tank. In this case we simplify by disregarding pipe friction, thermodynamic losses (like failure to fully expand due to cooling) and any other losses you can think of, and assume the potential energy of the compressed air is converted to kinetic energy. The potential energy PE = 0.5*P*vol (where vol is tank volume = L*pi*D^2/4). The kinetic energy of the spud is 0.5*M*v^2. Thus v=SQRT(P*vol/M).\nAbout other questions, 1 psi = 6894.75729 Pascals, and speed of sound =340.29 m/sec.\nNote that with what you have given in your question, you have a lot of units conversion to do.\nEDIT: See friction stuff added above. The spreadsheet includes friction effects.
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
1169918
1 ->25000\n2cm ->2*25000cm=50000cm=500m\n\nans: b:500m
481154
2 km = 2000m\n2000m = 200000cm
712308
speed of bullet=200mph\n=200*1.609343 km/h\n=321.8686 km/h\n=321,868.6 m/h\n=32,186,860cm/h  \nthe bullet has to be cut every \n1/10 th cm\nnumber of cuts/hour required\n=321,868,600\nthe fan has four blades,therefore,\nthe fan cuts four times per rev\nnumber of revs/hour required \n=321,868,600/4\nbut,there are 3600 secs in an hour\nnumber of revs/secs required\n=321,868,600/(4*3600)\n=22,351.98611\n\ntherefore,the blades of the fan have\nto make 22,352 revs/second \nin order to cut the cm long bullet \ninto 10 pieces\n\ni hope that this helps
869691
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)