What is the angle of banking in an aircraft and how it is calculated?
340987
An aircraft has 2 methods of turning. One is yaw, in which the rudder on the tail simply moves from side to side, causing excess drag on one side and hence turns the plane. Greater turning is achieved by first rolling the plane along its axis and then pitching upward the nose upward, which is now actually sideways.
2504
More than Bernoulli's principle and wing shape (airfoil) are at work in keeping a plane in the air. The main criteria is the  "angle of attack" of the wing. That is, the angle at which the wing meets the air passing by it. When a plane is flying, its wing is always at a positive angle of attack; the leading edge of the wing is higher relative to the air going over it than the trailing edge. So the air pushes the wing up.  When the plane is inverted, the angle of attack must be great enough to overcome the "negative lift"  created by the airfoil. This is accomplished by adjusting the elevator. Inverted flight will always require the input of down elevator.
1001145
The angle of elevation means that it's the angle that starts left to right and opens up. ( like <)\n\nTan 67 = opp/adj\n\nTan 67 = x/  50\n\nx = 50Tan 67\n\nwhatever that is
1147917
The geometric orientation of any planar geological feature (e.g., bedding planes, joints, or faults) is described in terms of the feature's strike and dip.  The strike is the compass direction of the line formed by the intersection of the planar feature with the Earth's surface, and the dip is the angle that the plane makes with respect to the surface.  \n\nIf you are making measurements in the field, one uses a compass and "clinometer" (a device that measures angles) to determine the strike and dip, respectively.  (by taking a compass bearing and measuring the dip angle).\n\nIf you are looking at a geologic map and trying to determine the strinke and dip of a feature, then this is indicated on the map by a special symbol that looks like a short line with another, shorter line on one side of the line at right angles to the first (i.e., a "tick mark").  There will be a number next to the shorter line.  The strike direction is given by the orientation of the longer line, and the dip direction by the shorter "tick mark".  The dip angle is given by the number next to the short "tick mark".\n\nSee sources.
499929
The first 4 is the maximum mean line camber in % of chord.\nThe second 4 is the position of the max camber in tenths of c.\nThe 12 is the maximum thickness of the airfoil in % c.\n\nThe airfoil you mention was derived by the folks at NACA in the late 1920s and was used on such planes as the Luscombe and  Aeronca.  Definitely a vintage model........
139495
Let's define some variables for the equation.\n\nD = monthly deposit ($200)\nr = annual interest rate (10% is 0.1)\nn = compound frequency (12 months in a year)\nA = account balance\ni = compounded interest\n\nLet's start the account this month, or month 0.\n\nA(0) = D\n\nOne month later, you make another deposit and earn some interest. But you only earn 1/12 of the 10% with you existing balance, which is r/12.  \n\nA(1) = D + A(0) + i(1)\n\nwhere, i(1) = A(0) * r/n. Which makes\n\nA(1) = D + A(0) + A(0) * r/n , or\nA(1) = D + A(0) * (1 + r/n ), or\nA(1) = D + A(0)*R, if R = 1 + r/n, but A(0) is D, so\nA(1) = D + D*R\n\nTwo months later, you make another deposit and earn some interest on the balance A(1). Similarly calculating,\n\nA(2) = D + A(1) * R, or\nA(2) = D + (D + D*R) * R, or\nA(2) = D + D*R + D*R^2\n\nKeep on going to three months later, \n\nA(3) = D + A(2) * R, or\nA(3) = D + D*R + D*R^2 + D*R^3, or\nA(3) = D * (1 + R + R^2 + R^3)\n\nFollowing this pattern, k months later, we get\n\nA(k) = D * (1 + R + R^2 + ... + R^k)\n\nNow, let's calculate the general expression (1 + R + R^2 + ... + R^k) in a closed form.\n\nLet S = (1 + R + R^2 + ... + R^k). Multiply R on both sides, we get\nS*R = (1 + R + R^2 + ... + R^k) * R, or\nS*R = R + R^2 + ... + R^k + R^(k + 1)\n\nSubtracting the two equations S*R and S on both sides, we get\n\nS*R - S = R + R^2 + ... + R^k + R^(k + 1)  - (1 + R + R^2 + ... + R^k)\n\nNotice the terms R all the way through R^k on the right hand side are cancelled and only left with R^(k + 1) - 1. So we have,\n\nS*R - S = S(R - 1) = R^(k + 1) - 1\n\nDividing (R - 1) on both side, we get\n\nS = [R^(k + 1) - 1] / (R - 1)\n\nRewriting the account balance at month k,\n\nA(k) = D * (1 + R + R^2 + ... + R^k), or\nA(k) = D*S = D * [R^(k + 1) - 1] / (R - 1)\n\nNow, 26 years make 26 * 12 = 312 months. But since you start with 0, so your k is actually 311. R = 1 + r/n = 1 + 0.1/12, and D = 200. So\n\nA(311) = 200 * [(1 + 0.1/12)^(311 + 1) - 1] / (1 + 0.1/12 - 1), or\nA(311) = 295667.15330119169 = 295667.15 (round to nearest penny)
184797
find the angle between two sides.
898116
A = b * h / 2\ndA/dT = [b * dh/AT + h * db/dT] / 2\n\nWhen h = 20 and A = 80, b = 8\n\n4 = [8 * dh/dT + 20 * 1] / 2\n4 = 4 * dh/dT + 10\n-6 = 4 * dh/dT\n-3/2 = dh/dT\n\nAs you can see, Gopal has the 4 * dh/dT part correct, but where Gopal has 8 minus 20, it should be 4 minus 10. As a result, Gopal's answer is twice the correct one. Gopal has a lot of good ideas, but is careless with their implementation. You can see that Gopal has unbalanced parentheses and brackets, no doubt the source of the confusion.\n\nAs a rough check, we can ask what the status is one minute after the time given in the problem, when the area is 84 and the base is 9. \n\nh = 2 * 84 / 9\n\nThe altitude computes to 18-2/3, much closer to my rate of change than to Gopal's.
832160
The angles formed by crossing lines are supplementary (add to 180º).\n\nSo it is a simple matter of subtracting:\n180º - 145º = 35º\n\nThe acute angles will each be 35º.
906923
Suppose the component vectors are a and b with magnitudes |a| and |b| and the angle between them θ\n\nThen, on completing the parallelogram of forces, the angle between the end of a and the beginning of b is (180° - θ)\n\nIf the resultant vector is R (which is the required vector), makes an angle of α between a and R, then by the sine rule\n\nsinα/|b| = sin(180° - θ)/|R|\nSo sinα = |b|sinθ/|R| (because sin(180° - θ) = sinθ\nHence α = arcsin{|b|sinθ/|R|}\n\nNote If you need |R| then by the cosine rule:\n\n|R|² = |a|² + |b|² - 2*|a|*|b|*cos(180° - θ)\n= |a|² + |b|² + 2*|a|*|b|*cosθ (since cos(180° - θ) = - cosθ)\n\nThis is for a and b in any orientation on a plane (not necessarily at right angles)\n\nIF θ = 90° then cosθ = 0 and sinθ = 1\nand these equations reduce to\n\n|R|² = |a|² + |b|²\nand \nα = arcsin{|b|/|R|} (which incidentally also equals arctan {|b|/|a|}
507693
I'll get you started...\n\nFirst, determine how long the ball was in the air.  From that, determine the velocity at which the ball was thrown (the rate necessary, factoring in gravity's acceleration, to keep it airborne for that period of time).  Construct a vector using the velocity of the truck vs. the velocity of the ball to determine the angle.\n\nGood luck!
38489
On an airplanes wing, the top is curved in shape, and the bottom is straight.  When the plane gains speed, the air will be forced to move faster over the curved portion.  The faster moving air applies less pressure than over the straight portion therefore creating "lift"
74297
(# sides - 2) * 180 = sum of all the interior angles\n\n[(# sides - 2) * 180]/(# sides) = each interior angle if they're all congruent
772033
Draw a picture of it with arrows representing the vertical and horizontal velocities as 2 legs of a right triangle.  This triangle will show you that the angle with the vertical is arctan 3.5/5.5 = theta\n\nThen the flight path V will be Vv/cos(theta) or Vh/sin(theta)\n\nSince you're ruling out pythagoras, you really need to find the angle first, then the speed.......
907874
You should define your terms:\nWhat is µ?\nWhat is ¹s?\n\nAssuming µ is the angle the plane makes with the horizontal (traditionally this is Θ), and assuming ¹s is the coefficient of static friction between the block and the plane (traditionally, this is µ):\n\nI - T\nII - F\nIII - F\nIV - F\nV - F \nVI - T\n\nAnswer choice (2) satisfies these requirements.\n\nI always hated these types of questions.  They test your bookkeeping skills moreso than your knowledge of physics.  A simple T/F quiz is always better.
1187062
a) the 90 degree angle
602899
I always find it easier if you draw it first, which this stupid bos wont let me do.\n\nYou know one angle (theta) and one side of the triangle on parts a and b.  Use Tan Theta = x / 30 to find how far the truck has traveled in one second, and convert to MPH.\n\nFor C, you have the distance traveled in MPH and one side (30 ft).  Convert 55 MPH to ft / sec and find the angle Theta.  Tan Theta = 55 MPH / 30 ft\n\nYou are going to have to figure the answers out for yourself.  Good luck