what is the equation of drag coeficient?
151103
The drag coefficient is a number that aerodynamicists use to model all of the complex dependencies of shape, inclination, and flow conditions on aircraft drag. This equation is simply a rearrangement of the drag equation where we solve for the drag coefficient in terms of the other variables. The drag coefficient Cd is equal to the drag D divided by the quantity: density r times half the velocity V squared times the reference area A. \n\nCd = D / (A * .5 * r * V^2) \n\nThe quantity one half the density times the velocity squared is called the dynamic pressure q. So \n\nCd = D / (q * A) \n\nThe drag coefficient then expresses the ratio of the drag force to the force produced by the dynamic pressure times the area. \n\nThis equation gives us a way to determine a value for the drag coefficient. In a controlled environment (wind tunnel) we can set the velocity, density, and area and measure the drag produced. Through division we arrive at a value for the drag coefficient. As pointed out on the drag equation slide, the choice of reference area (wing area, frontal area, surface area, ...) will affect the actual numerical value of the drag coefficient that is calculated. When reporting drag coefficient values, it is important to specify the reference area that is used to determine the coefficient. We can predict the drag that will be produced under a different set of velocity, density (altitude), and area conditions using the drag equation.\n\nThe drag coefficient contains not only the complex dependencies of object shape and inclination, but also the effects of air viscosity and compressibility. To correctly use the drag coefficient, we must be sure that the viscosity and compressibility effects are the same between our measured case and the predicted case. Otherwise, the prediction will be inaccurate. For very low speeds (< 200 mph) the compressibility effects are negligible. At higher speeds, it becomes important to match Mach numbers between the two cases. Mach number is the ratio of the velocity to the speed of sound. At supersonic speeds, shock waves will be present in the flow field and we must be sure to account for the wave drag in the drag coefficient. So it is completely incorrect to measure a drag coefficient at some low speed (say 200 mph) and apply that drag coefficient at twice the speed of sound (approximately 1,400 mph, Mach = 2.0). It is even more important to match air viscosity effects. The important matching parameter for viscosity is the Reynolds number that expresses the ratio of inertial forces to viscous forces. In our discussions on the sources of drag, recall that skin friction drag depends directly on the viscous interaction of the object and the flow. If the Reynolds number of the experiment and flight are close, then we properly model the effects of the viscous forces relative to the inertial forces. If they are very different, we do not correctly model the physics of the real problem and will predict an incorrect drag.\n\nThe drag coefficient equation will apply to any object if we properly match flow conditions. If we are considering an aircraft, we can think of the drag coefficient as being composed of two main components; a basic drag coefficient which includes the effects of skin friction and shape (form), and an additional drag coefficient related to the lift of the aircraft. This additional source of drag is called the induced drag and it is produced at the wing tips due to aircraft lift. Because of pressure differences above and below the wing, the air on the bottom of the wing is drawn onto the top near the wing tips. This creates a swirling flow which changes the effective angle of attack along the wing and "induces" a drag on the wing. The induced drag coefficient Cdi is equal to the square of the lift coefficient Cl divided by the quantity: pi (3.14159) times the aspect ratio AR times an efficiency factor e. \n\nCdi = (Cl^2) / (pi * AR * e) \n\nThe aspect ratio is the square of t
728987
for lower speeds, i.e. not near speed of light\n\nke = 1/2 m v^2\n\nSo doubling speed = quadruple energy.\n\nAdditionally drag is also proportional to the square of velocity so that the amount of power necessary to push a car is proportional to the cube of the velocity.
985162
This is a problem in open channel flow in Fluid Mechanics, not a pre-calc problem.  A proper solution needs calculus and a better description of the harbor bottom.\n\nAn intuitive set-up based on the information you've given would be \nh = (e^-kt)[(10 - 9) - 9sin(2πt/15)]\nwhere h is in feet, t is in minutes, and k is a damping constant, probably on the order of 1/15.
412953
An equation that is incorrect.
1165015
The diffential flow rate betwwen air and fluid depends on the following factors. An exact equation is very difficult.\n1. type of fluid used because of the viscosity of the fluid.\n2. altitude and barometric pressure of the test location.\n3. test pressure\n4. temperature of the test air.\n5. many more\n\nAs you can see the formula for this would be very long. A rough formula you may try would take into account.\n1. volume of air at what rate.  in2/sec\n2. pressure for the air to obtain that volume\n3. viscosity of the fluid\n4. use 1 and 2 as baseline with air density\n5. compare 4 with fluid viscosity at desired pressure\n\nI know I couldn't give you a formula but I hope this may help you.\nYou may try math2.org for formulas or search for math formulas.
974937
please complete the equation
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
961494
This is an energy balance situation.  The work done in dragging the object must equal the energy change.  In this case the energy is potential energy, given by m*g*∆h, where ∆h is the change in vertical distance.  The work (energy) required to drag the block is force times distance along the incline.  Equate these two quantities and solve for F.
742135
There are several forces at work here.\n\nHorizontal: drag due to air friction slowing the horizontal velocity down\nVertical: weight, speeding the vertical velocity up\n\nThe total velocity when breaking surface tension is the vector sum of the horizontal (vh) and vertical (vv) velocities = V = sqrt(vh^2 + vv^2).  The vertical velocity = vv = gt = g sqrt(2S/g) = sqrt(2Sg); where S = 60 ft and g = 32.2 ft/sec^2.\n\nWe can't find vh with what you've given us because the initial horizontal velocity vh0 = 150 mph will be slowed down by the drag friction.  And drag friction forces depend on air density, coefficient of drag, and the cross sectional area of the person...none of which has been given.\n\nHowever, one rule of thumb in such matters is that the terminal velocity of a person falling through the atmosphere is about 120 mph.  So we can suppose that vh1, the end horizontal velocity, will be somewhere between 120 and 150 mph.  Let's assume the soft landing and vh1 = 120 mph = 176 ft/sec as the terminal horizontal velocity.\n\nThus, the break surface tension velocity V = sqrt(vh^2 + vv^2) = sqrt(176^2 + (2Sg)^2) = sqrt(31000 + 3800) = 186 ft/sec or about 127 mph.  Ouch.  At those speeds, water tends to act like concrete...in other words, splat...end of XXX.
566230
Set up the equation\n\nP = dRT\n\nAssuming P to be constant,\n\ndT = d'T'\n\nor d' = 0.76*273/298 = 0.696 g/l.
776431
Integrate the acceleration equation with respect to time, twice.  Use, Xo = 500 and Vo = 10...solve for t.
108206
I think you can use Bernoulli's equation to predict the difference in pressure from one area of the hose to the other and, therefore, the pressure drop:\n\nP = (flow rate * 8 * viscosity * length of hose) / (pi * hose radius^4)\n\nNot sure if it works though...typically used for fluids...
843122
good start.\n\ntotal inital momentum = 480-397.5\n\nthis is an inelasitc collision, and aferwards you have \nboth masses travelling at velocity v:\n\n(75+120)v = 480-397.5\n\nsolve for v.  This is the velocity at which they both slide.\nThe inital energy now is 1/2(75+120)v^2\n\nThis energy is converted to frictional energy:\n\ninital kinetic energy = total work of friction = Fr(distance)\n\nFr = uN = u(120+75)g = .3(195)(9.8)\n\nso we have\n\n.3(195)(9.8)(distance slided) = 1/2(195)v^2 \n\nYou know v form above.  So find distance slided.
778679
I believe you will find the solution to your problem here:\n\nhttp://www.engineeringtoolbox.com/bernouilli-equation-d_183.html
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
748998
Assume smaller pipe fills the tank in x hours and the larger one in (x-2.5) hours. Taking the given inf that both together can fill the tank in 8.4 hours, develop a quadratic equation in x, and solve it. \n\nLarger pipe fills the tank in 15.65 hours whereas the smaller one takes 18.15 hours to fill it.