an aircraft loops the loopof radius R=500m with a constant velocity v=360km/hr see below?
612424
I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
758142
a) There is an equation derived from the motion in x and y that results like:\n\ny = (tanθ )x - g x²/ [2(Vo cosθ)² ] we know y, x and θ we need to solve for Vo:\n\nVo =  (x / cosθ) √[ g / (2 (x tanθ  - y )]\nIf we put  the plane at the initial hit at 1 m, y is 21 m -1m = 20 m\nAnd as we know that x = 130 m and θ= 35°\nSubstituting:\n\nVo = 41.7 m/s <<<<<<<<\n\nb) t = x / ( Vo cosθ )  Substituting: t = 3.81 s <<<<<<<\n\nc) Vfy = Vo senθ - gt   (Vfy is the final velocity of y after the time t)\nSubstituting, its components are:\nVfy = -13.46 m/s      <<<\nVox = Vfx = Vo cosθ = 34.16 m/s   <<<\n\nand its magnitude:\n\nVf = √(Vfx² + Vfy²) = 36.71 m/s <<<<<
889727
To find the tangential velocity, we can use the formula\n\nV= r w\n\nwhere r is the radius and w is the angular velocity. Since we know r all we need to do is convert w to rad/sec to find V.\n\nw = (50,000 rev/min)(2 Pi rad/rev)(1 min)/(60 sec)\n= 5235.99 rad/sec\n\nNow we have everything needed to find V.\n\nV = (.25 m)(5235.99 rad/sec)\n= 1309 m/s\n\nSo the tangential velocity is 1.3 km/s\n\nThe energy is stored in the kinetic energy of its rotation. The formula for this energy is\n\nK = (1/2) I w^2\n\nwhere I is the moment of inertia and w is the angular velocity. From the description of this flywheel, it sound like it is a loop rotating about a central axis. The rotational inertia of this shape is\n\nI = M R^2\n= (50 kg) (0.25 m)^2\n= 3.125 kg m^2\n\nUsing our previous calculation we know w in rad/sec.\n\nSo now we have all the information we need.\n\nK = (1/2)(3.125 kg m^2)(5235.99 rad/sec)^2\n= 4.3 x 10^7 J = 43 MJ\n\nThe amount of work that could be done by the flywheel assuming that all energy could be converted. The amount of work is the difference in the kinetic energy. Assuming all of the energy could be converted, this would be the total amount of work.\n\nW = Kf - Ki\n\nwhere Kf is the final kinetic energy and Ki is the initial kinetic energy.\n\nKf = 0\nKi = 43MJ\n\nW = (0) - (43MJ)\n= -43MJ\n\nIt should be noted that this is the amount of work required to stop the rotation of the flywheel, so the absolute value of this is the total amount of work that the flywheel can perform.\n\nW = |-43MJ|\n= 43MJ\n\nIf we can not get any work below 10000 rev/min, we can use the same formula. We just need to find out K for that speed. Recall that\n\nK = (1/2) I w^2\n\nI is still the same, so all we need to do now is change w to rad/sec.\n\nw = (10000 rev/min)(2 Pi rad/sec)(1 min)/(60 sec)\n= 1047.2 rad/sec\n\nSo now we have the information to find Kf\n\nKf = (1/2)(3.125 kg m^2)(1047.2 rad/sec)^2\n= 1.7 x 10^6 J = 1.7 MJ\n\nSo now we can find W\n\nW = |Kf - Ki|\n= | (1.7MJ) - (43MJ) |\n= 41MJ\n\nIf we are only able to get 75% of this work, then we must multiply this number by 75%\n\nWact. = (.75)(41MJ)\n= 31MJ\n\nwhere Wact is the actual work. The rest of the energy goes to friction of the wheel spinning on it's axis, probably on bearings or whatnot, to friction with the air around it, imperfections in the disc also cause it to be off balance and wobble, loosing energy. It goes lots of places. But friction is probably the answer you're looking for.\n\nI hope this is right, it's been a while since my physics class. The order of the questions seems to imply that you were supposed to use the tangential velocity to figure out the amount of energy, but I thought this way was easier for me. You can, and should, still try it that way to make sure the answers are the same. If they are not then one of us made a mistake... and I wouldn't be surprised if it was me.
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
883859
energy dissipated in the region AB is mu*m*g*d=3.75J\nenergy stored in spring = 0.5 *k*x^2  = 0.5 m*(vb)^2 + energy dissipated in region AB   -(1)\nwhere vb is the velocity of the block at the bottom of the loop\nvb>=sqrt( 5*g*r) for completing the circle vetically\nNOTE: velocity at the top of the loop is atleast sqrt(g*r) so that the centrifugal force balances mg and the block doesnt lose contact\nfrom ---(1) : 0.5 kx^2 >= 0.5* m * 5*g*r +  7.5\nso x = 0.5357 metres
1133529
Approximately 1,038 miles per hour. Actually, slightly more depending on the plane's altitude. At 6 miles altitude above the earth, it would add another 2 * pi * (6) / 24 miles an hour, or 1.57 miles an hour, to the required speed.
767928
1. its final velocity depends on the direction it was heading to start off with. If it was flying in the x direction (no components in y or z) then after 4.3 seconds the x component has increased by 8.92*4.3  and the y component has increased (from 0 this time) by 1.21*4.3.\nso now you can work out the new direction (in the x direction but slightly downwards by whatever angle has a tangent of 1.21*4.3/(152+8.922*4.3) \nand using pythagoras to determine the speed.\n\n2. You can ignore the gravity as it is << the precision with which the acceleration due to the electric field is stated (I'm assuming the 106 is 10^6 = 1000000) \nso the final velocity is 3.80*2.45*1000000 m/s upwards.
874378
If you mean air speed - it is measured in knots.\nIf you mean altitude it is measured in metres.\nIf you mean distance flown - kilometres
724756
For a body in circular motion there will be a centripetal force\n= M V V /R.\n\nIn this case the tension of the string provides this force.\n\nMaxinmum tension = 50N.\n\nThat is the maximum centripetal force is 50 N\n \nM V V /R = 50\n\nV V = 50 x R / M = 50 x 1/ 2 =25 (m/s)^2.\n\n The maximum speed is V = 5 m/s.
1069494
It will be more. Mathematically:\n\nno wind case:\n\nLet the distance from A to B = d\nAnd let the speed of the plane without wind = v\nSince distance/speed = time, in the case with no wind the time for the return trip =  d/v + d/v = 2d/v\n\nWind case:\n\nLet d,v be as above, and let the wind speed blowing from A to B = w. The planes wind-aided speed = v+w and it's wind reduced speed is v-w. The time for the return trip is now:\n\nd /(v+w) + d /(v-w) = (d(v-w)+d(v+w))/(v^2 - w^2)\n\nThis simplifies to:\n\n2dv /(v^2 - w^2) compared to 2d /v for the no wind case.\n\nThe ratio of the no wind case to the wind case is:\n\nv^2 / (v^2 - w^2)\n\nWe see that if w > 0, this ratio will be greater than 1, which means the return trip will take more time if there is a wind.
958243
You need to know the initial speed.  Then the answer is:\n\n(v - u)/t\n\nwhere u = initial speed\n           v = 0 (final velocity)\n           t = 5 seconds\n\nThe answer will be negative because it is decelerating
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
396138
a=F/m, lim x->inf x/x = 1. The object will accelerate.
925006
Since the magnitude of the acceleration is v^2/r, and the radius of the Earth is approximately 4000 miles (21120000 ft), you would have to solve the following equation: v^2/21120000=32\nv^2=675840000\nv=25997feet per second  or\nv=4.92 miles per second
890224
Velocities are vectors, so you´ll have to proceed according to vector rules for addition/subtraction.  Draw a horizontal line, with an arrowhead pointing to the left. This represents the (unknown) velocity due West. Now draw another line, vertical, this time.  Put an arrowhead in the lower end of this line.  This stands for the 50 kph wind speed to the South.  These two lines should make a right (90°) angle.  The lines haven' t to be drawn to scale.\n\nFinally, draw an oblique line connecting both remaining ends.  This is the aircraft speed; to indicate direction, put an arrowhead  at its leftmost end.\n\nNow, you have a triangle of which two sides are known.  This, and the fact it is a right triangle, suffices to solve the entire triangle. The angle you seek is that formed between the horizontal and oblique lines.  Let θ denote this angle.  From elementary trigonomety, sin θ = opposite side/hypotenuse = 50/205 = 0.2439... All you have to do now is to find the angle whose sine is 0.2439...  This is 14.1169...°.  From the figure you have drawn, this angle lies N of W.\n\nThere are several ways to figure out the speed to ground.  You may use Pythagoras' Theorem; or, having found the opposite angle, you may compute it as 205 × cos 14.117° = 198.809...\n\nJust follow the same procedure to solve the other problem.  You'll have to decide which direction is "river across" and which is "downstream"; I´ll suggest "right" and "down", respectively.  Just plug in the proper figures.