what is the stick in cocpit and how does it work  on which speed normally aeroplane takes off .thankyou?
535253
The stick in the cockpit allows for the control of the aircraft. i.e. ailerons, elevator. In the old days, the stick was physically attached to wires that attached to the control surfaces such as the ailerons. Nowadays, the stick actuates computer controlled sensors which then actuate servos and control mechanisms. The take off speed depends on what type of plane it is and what speeds the plane is designed for. The take off velocity is normally 1.2 x the stall speed...which is the absolute minimum speed the plane will fly without crashing. For a "rough idea" of take off speed...150 knots or so is a good number.
47938
The first answer is correct. I think for the second answer, the questioner would probably be wondering...well, if the helicopter takes off with the same velocity as the earth, since the heli is now hovering and the earth keeps on turning, why doesn't the earth spin past the helicopter...\n\nAnd that is simply because the atmosphere is turning around with the heli, like the first answer stated.\n\nJust wanted to elaborate with another example. A fly takes off from the dashboard of a speeding car with its windows closed. Does the fly stay in air or, since the fly is now seperated from the car, does the car rush forward, making it seem like the fly is hurtled to the back of the car?\n\nEven in this case, the air inside the car is being brought along with the car at the same speed. Therefore, the speed of the fly relative to the car will always be 0.
791539
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²
483369
Runway 21 would be into the wind, but you're going to be applying right rudder to counter the left yaw disturbance due to the left engine out, and will be sideslipping to the left. To keep you on course you need the headwind to be at least partially a crosswind from the left, so given the choices, that would be runway 30.\nAnother reason to have the crosswind from the left: If you are going to use reverse thrust to decelerate, "Given a choice, it is advantageous to put the cross wind on the same side as the failed engine." (ref.)
997128
Actually yes\n\na small aircraft, can in fact stop forward movement over the ground.  If the plane is flying into the wind at 70 mph, (indicated air speed) however the wind  he is flying into, is hititting his plane at 70 mph, that would put the ground speed of zero.  All aircraft are effected by wind.  The same plane flying at a air speed of 70 mph, with a tail wind of 70 mph, has a ground speed of 140 mph.\nalso\nA landing is also known as a controlled crash.....think about it
1005590
hi buddy!\nscience has not developed so much that planes could fly at a speed equal to light!!!!\nhowever, if a plane is so developed, you will be able to see the light if you switch it on inside the aircraft, the same theory is applied for super sonic aircrafts. if there are two pilots inside the aircraft, they can hear their talk as both are flying at the speed greater than sound. here, relative speed comes into play. airsraft and both pilots acts as a single body. so, relatively, they are all still to each other. so, they can easily talk to eachother.\n\nif you turn on the switsh which would light a bulb outside the aircraft, then you will not be able to see it. same happens in super sonic planes. if a sound is made outside the aircraft, the pilots will never be able to hear it.!!!\n\nhope that i have answered ur querry!\nif further clarification is needed, u can mail me at mangeshome1982@yahoo.co.in
90020
Every object touching and having good grip on the body of the train, move along with the train.This means that every object inside the train moves with the same velocity as that of the train.Accordingly, fly is also moving with the train. It has gained this speed by sitting on the interior body of the train.According to theory of physics, absolute velocity has no effect and diffrence when the frame of reference is within the object.Therefore fly while flying makes two kinds of motion 1) It is moving along with the train 2) it`s flying in its desired directions.The motions (1) & (2) are true to the observer outside the train.But, motion (2) is true to the observer inside the train because the observer inside the train is also moving along with the train.Moreover, When the fly tries to enter the fast moving train from outside the train, then the probability of striking on the panel (wall) of windows is more with respect to the speed of the train.This is because the fly is not in the speed of the train.You might have observed, some of the men on the platform enter the moving train by moving along with the train for some distance and jump into the train. I think this is enough to comprehend.
335082
sorry cant help you. don't knopw the answer. consult your sky manual for uinformation on control problem
561298
u will like this one http://www.acepilots.com/planes/main.html
503553
What a great question!  I understand the reasons for the inversion, but the question is why always left to right (a clockwise direction).\n\nI'm guessing it has something to do with the convention that the pilot always sits on the left in the cockpit.  (Hey, if this were a British space shuttle, would the pilot sit on the right?  Just kidding, of course, I know that pilots in British planes follow the same conventions. :-)  )
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.
262603
it all comes down to the difference between laminar flow and turbulent flow.  think about a non-spinning ball moving through the air:\n\nif the surface of the ball is perfectly smooth, then the motion of the air just near the surface (this is called the boundary layer) will be perfectly smooth too.  as the air moves around the sides towards the back of the ball, the angle between the direction of the air's motion and the surface of the ball will get bigger and bigger.  \n\nat some point the boundary layer won't be able to "stick" to the surface anymore and will detach.  (this phenomenon is called boundary layer separation.)  the faster the air is flowing, the separation point will be closer to the place at which the ball is widest.  this separation means that, behind the separation point, the air near the ball's surface will be turbulent.  it is essentially what happens to an airplane's wing during a stall.  \n\nbecause the ball is symmetric, the turbulent area will be equally spread between the left and right sides of the ball.\n\n(a brief aside: laminar flow exerts a greater pressure on the ball's surface than turbulent flow does.  this is why aircraft engineers go to great lengths to make the wings of a plane very smooth - if the airflow around the wing is very smooth too, then the wing can generate greater lift.)\n\nanyway, so everything's symmetric for a ball that's not spinning.  now think about a ball that's spinning around it's vertical axis.  the relative velocity of the air to the ball's surface is different on the left and right side because of the spinning.  this velocity difference means that the boundary layer separation happens at a different place on the left or right side.  \n\nthis means that the left/right side of the ball will have a larger/smaller area with turbulent flow at the surface.  the symmetry is broken and the resulting pressure difference causes a net left/right force on the ball, which makes it curve.\n\nthere is a nice picture of the non-spinning ball at the bottom of this page:\nhttp://astron.berkeley.edu/~jrg/ay202/node97.html\n\nhere's a paper about it, with a nice picture that shows the laminar and turbulent asymmetry:\nhttp://www.wooster.edu/physics/JrIS/Files/nowicki.pdf\n\nmore pictures near the bottom of this page:\nhttp://www.aerospaceweb.org/question/aerodynamics/q0215.shtml\n\ncheers!
962720
Lots of unnecessary complications here.\n\nAirspeed is the hypotenuse. 20.4 is the side opposite the angle we are looking for.\n\nThe heading, east of north, must be at angle sin^-1(20.4/241).\n\nSpeed relative to the ground is the third side of the triangle, and is \n\nsqrt(241^2 - 20.4^2)\n\nZidane has got the angles all wrong. The wind is from the east. We want to fly due north. That's where the right angle is. 241 is the hypotenuse of the triangle. The wind serves to make the ground speed less than 241, not greater.
463967
The first answer gives ground speed.\n\nAir speed is measured by a differenctial pressure arrangement, in which the air pressure on the leading edge of a wing is compared with the "static" pressure (a reading which is shielded from the effect of the aircraft's movement through the air.)
14321
First, look at any typical airfoil shape. It is sort of teardrop shaped, with a round end and a pointy end, but not symmetrical--one side seems flatter than the other. Let's imagine that the pointy end points to the right, the round end point to the left, the flatter side is on the bottom and the rounder side is on top.\n\nLet air flow from the left to the right. Air flowing near the airfoil at speeds that are not too fast (less than the speed of sound) is not compressible. Let's follow two tiny blobs of air that start next to each other, but approach the round end of the airfoil in such a way that one blob goes up and over the airfoil, while the other slides over the bottom. These two blobs will arrive at the pointy end at approximately the same time. But, the one going up and over as a longer path to take and therefore must travel faster. All the air molecules in a blob are actually thrashing about randomly and can collide with other air blobs nearby as well as the upper surface of the airfoil. This is true of the upper blob and the lower blob. But, if the upper blobs average speed is faster, then its molecules collide with the upper surface of the airfoil less over the trip from round end to point end than do the molecules in the lower blob. Therefore the pressure caused by the collisions of the molecules in the upper air blob on the upper side of the airfoil is less than the pressure on the lower surface caused by collisions of the lower blob's molecules with the lower surface. This means there is a difference of pressure between the two surfaces--pressure drops as you go from upper to lower surface. If the airfoil is rigid, then this difference in pressure will become a force exerted on the airfoil from below, called lift. This lift is what makes an airplane rise on takeoff as it rushes through the air with its wings whose cross-sections are airfoil shaped, rounded end into the oncoming air and flatter side towards the ground.
899525
Using the example....from the problem you can tell that the tail wind helps the plane move faster.  Let's say the speed of the plane in still air is x, so the speed of the plane with the tail wind is equal to x+25.  The distance the plane travels is equal to (x+25)5, which is the speed of the plane times the number of hours the plane flew.  Now for the return flight, we know the wind is now working against the plane, so it reduces the plane's speed.  Therefore the speed on the return flight is x-25.  The distance traveled can be calculated by (x-25)6.  Since the flight to the destination and the return flight must be the equal, we can set the  distances equal to each other.\n\n(x+25)5 = (x-25)6\n5x + 125 = 6x - 150\n\nFrom there you can solve the problem.
951198
^ Parasite alert LOL! ^\n\n30 s for C to move from A to B, maybe? I'll assume that. Also, x seems to be a length, not an angle. Maybe a) is asking for the rate of change of z (called z')? We know the angular rate of C about O (y') is pi/30 (answer to b)), and triangle CAO is isosceles, so z'=-0.5*pi/30. \nFor c), circumferential velocity is pi*r/30. Angle CAO is, conveniently, 45 deg, so x' = sqrt(0.5)*circumferential velocity.\nFor d), the area of triangle OAC is .5*AO*altitude, and altitude=r*sin(y). This is a maximum when y (not x) =pi/2.\nSorry, I know you wanted  no answers, but I couldn't see how/where to stop short of most of the answers.