what is L, the distance traveled along the incline by the block after it exits the gun?
927920
The distance is V²/2a where V = muzzle velocity and a = g(sinθ + μcosθ) where θ = angle above hoizontal of the incline and μ = coefficient of friction.\n\nNote that when θ is <0, a can be 0 and the distance is infinite, as it never slows down at all.
703031
Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
944006
a) Initial velocity, u , of the bullet as it hits target=1200m/s\n    final velocity, v, as it emerges =850m/s\n    distance traversed,s = thickness of target\n                                   =30cm\n                                   =0.30m\n\nv^2-u^2=2as\n\n850^2-1200^2=2a*0.30\n722500-1440000=0.6a\n-717500=0.6a\na=-717500/.6\n  =-1,195,833m/s^2\n\nv=u+at\n\n850=1200+(-1,195,833)t\n1,195,833t=1200-850\n               t=350/1,195,833    \n                =2.93*10^-4  s\n\n              \nb) F=ma\n mass of bullet, m=25grams or 0.025kg\n\nF=0.025*1,195,833\n  =29,896N\n\nc)Initial momentum of bullet=25*1200\n                                         =30,000\nInitial momentum of target=350*0\n                                          =0\n\nFinal momentum of bullet=25*850\n                                         =21,250\nFinal momentum of target=350v\n\nBased on conservation of momentum:\n\nInitial momentum of the system=Final momentum of the system\n\nSubstitute known values:\n\n30,000+0=21,250+350v\n350v=30,000-21,250\nv=8750/350\n  =25m/s
833335
Int(sqr(2y))\n\nIf you double the blocks, that gives you an 'almost' square since the diagonal of the incline made by the blocks would be 45 deg.  So take the square root of that to get an estimate of the number of rows (steps).  Take the integer of that number to get actual total steps.\n\nint(sqr(2*1)) = 1\nint(sqr(2*2)) = 2\nint(sqr(2*3)) = 2\nint(sqr(2*4)) = 2\nint(sqr(2*5)) = 3\nint(sqr(2*6)) = 3\n\nthis gives a close answer but apparently not full proof.  if you follow this idea further you may find a better equation.\n\nThe problem is when you double the number of blocks, you end up with extra row of blocks beyond a true square.\n\nInt(sqr(2y - Int(sqr(2y))))\n\n1 = 1\n2 = 1\n3 = 2\n4 = 2\n5 = 2\n6 = 3\n\nso this second equation looks good.  basically, I'm subtracting out the extra row.  I don't know if you are allowed to use the Integer function in your calculation but works good for programing.\n\n         1\n        11\n       111\n      1111\n     11111\n    111111\n   1111111\n  11111111\n 111111111\n1111111111
303961
let d=distance traveled in one direction(away from town or back)\ntime=distance/rate\n8hr= d/9 + d/3\nd= 18miles = answer
1135263
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.
10661
Depends on weight of the block angle of inclination, friction. Get resultant velocity by getting velocity along X (w cos theta) and velocity along y (w sin theta).... blahblahblah... check other sources but basically it goes like that.
104839
When an object is launched under the influence of only a uniform gravitational force field, it is said to follow a projectile motion. Under ideal condition, the path it traces is parabolic. There is catch here – if the initial velocity is equal or more than the escape velocity then the object will simply vanish into space i.e. leave earth!\n\nImagine if the launch angle from the horizontal is “theta” and initial velocity is “U” then you can easily compute vertical (Uv) and horizontal (Uh) components of it. The Uv will be equal to U x Sin(theta) and Uh will be equal to U x Cos(theta). Now by using Uv you can compute the max height it will reach and the total time it will take to go up and return to ground. With this round trip time in hand, you can easily compute the max horizontal distance it will cover before hitting the ground.\n\nAll these calculation assume ideal conditions like no wind resistance etc.
265191
Work = Force x Distance\n\nIn this case, since the shot is going up, i.e. opposite the force, the work done on the shot is negative.  Another way to look at it, if up is our positive direction, the distance will be positive but the force, i.e. the weight, is negative.  Note that the distance over which the weight acts is the difference in height, not the entire height of the shot relative to the ground.\n\nW = Force x distance = Weight x distance = - 71.6 N (2.09 m- 1.60m)\n\nI'll leave the arithmetic to you.\n\nTo answer the second part, just notice that the change in potential energy of the shot will be equal in amount to the work done on it.  Just take away the negative sign, since its potential energy is increasing. Another way of looking at it is to say that the work done on the shot takes away its kinetic energy, hence the negative sign, and converts it to potential energy.\n\nSince you're only concerned with the change in potential energy, you only have to concern yourself with the difference in height.  If you want to use the ground as the reference level, and calculate potential energy, you get\n\nDEL PE = PFf - PEi = Weight x final height - weight x initial height\n\nDel PE = 71.6 N (2.09 m- 1.60m)\n\nSo it works out to the same thing.  Yippee.
365737
Assuming you fire parallel to the ground, then the bullet will be at its maximum horizontal velocity when it leaves the barrel, and will be at minimum vertical velocity.  The bullet cannot increase in horizontal speed once it has left the barrel, as there is nothing to make it accelerate.\n\nDue to air resistance, the horizontal velocity will decrease as the bullet travels further and further.  The vertical velocity will increase, as the gravity of the Earth "attracts" the bullet.
550744
If I'm reading your description right, I think your first boat is heading up and to the right ("direction N.68 degrees E") making an angle of 68 degrees with north, or 22 degrees with east.  And I think your second boat is headed directly east ("S.90 degrees E").  So I'm assuming that they form a triangle, with one angle being 22 degrees and another one being a 90 degree angle, since we're looking for the point where one boat will be directly south of the other.\n\nThe first boat traveled 12t miles so far.  The second boat left an hour later and traveled 16(t-1) miles.\n\nThe first boat's path turns out to be the hypotenuse of the triangle, and the second boat's path will be one of the adjacent legs to the 22 degree angle.  So we can set up an equation with trig identities:\n\ncos 22 = 16(t-1)/12t\n0.9272 = 4(t-1)/3t\n2.7816t = 4t - 4\n4 = 1.2184t\nt = 3.3 hours\n\nSo after 3.3 hours, one boat will be directly south of the other.\n\nFinding that distance can be done one of several ways: using the Pythagorean theorem, sine relationships, law of cosines, etc.  I'll use sine, since we're already doing trig.  After 3.3 hours, the northernmost boat will have traveled 39.6 miles.  The distance between the boats becomes our side opposite the 22 degree angle.  So,\n\nsin 22 = distance between/39.6\ndistance = 39.6sin 22\n= 14.83 miles\n\nNow if I read the angle wrong, then simply go back and change all the "22" angles to the correct one and recalculate.
709316
assuming second block is originally at rest\n\nconservation of momentum says\n\nv11m1=v2m2+v12m1  (v11 initial block1 velocity, v12 final block 1 velocity)\n\nnow 1/2 mv11^2 = 1/2m2v2^2+1/2m1v12^2\nor m1v11^2 = m2v2^2+m1v12^2\n\n2 equations and 2 unknowns
189082
If this isn't a math question and you don't want to buy a tool and you don't have a protractor, there should be a leveled board on the ground, and your tilted board should be long enough to reach it. Mark the ground board where the tilted board touches it. Mark a point on the tilted board (the higher it is the more accurate your answer will be), drop a plumb line from there to the ground board, and mark where it touches. Measure the length from the bottom of the tilted board to the upper plumb-line mark (call it "A") and the distance between the two marks on the ground board (call that "B"). The cosine of the angle you want is B/A. Many calculators will give you the angle (using an arc-cosine function).
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
821275
4) Independent with one another...\n\nAs the projectile ascend/descend, its velocity has two component.. The vertical and horizontal. Remember that velocity is a vector quantity.\nThe vertical component detemines the maximum height the projectile will attain and also the time of its travel. The horizontal component determines the horizontal distance.
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec
1165042
the perfect distance is 25 yards.  this will give you an idea on how much drop your gun has.  at 50 yards you can dicover if you are pulling.