What is maximum terminal velocity in mph?
964449
Depends on what mass, what size due to air resistance\nA ball bearing coming from orbit could be in excess of 5mps, a man falling (not from orbit! but maybe 50,000') about 180mph.
604528
Gravity won't let you. You have to go into orbit first, and 1 mph will not do that. You have reach escape velocity to escape earth's gravity, which is 11.2 km/s.
977823
A.momentum is velocity times mass. Here this is 30*30'000=900'000 kg*m/s\n\nB.multiply m/s by 3.6 to get km/h, you have 30*3.6=108km/h\n\nC.distance = velocity * time, here this is 30 * 0.75 = 22.5 meters\n\nD.60 km/h divided by 3.6 gives you 16.67 m/s. Over 0.75s the driver would have travelled just 12.5 meters, i.e. 10 meters less.  So of course it is safer to driver more slowly.  But the catch is, with this reasoning you can show that it is even safer to drive at 40km/h, then 20km/h, and in the end you'll find that remaining at rest is really safe.  But at the end of the day, you still want the truck to get the stuff it transports somewhere, and if possible within a reasonable amount of time... ;-)\n\nhope this helps
806280
There has been one careless conversion of 50 mph to not quite the correct number of ft/sec.  Remember that 60 mph = 88 ft/sec exactly.\n\nS = V^2 / 2A, and V = 88 * 50 / 60, so S = 74.691. . . feet.
287006
Terminal Velocity.  It depends on the weight of the object because it is based on gravity.  If there is force applied to the object then it could travel faster for moments then gravity would slow it down.
985855
velocity = df(t)/dt = 4π + 4πcos(4πt)\nacceleration = dv(t)/dt = -16π^2sin(4πt)\n\nParticle is at rest when velocity = 0, or 4π+4πcos(4πt) = 0\n1+cos(4πt) = 0\ncos(4πt) = -1   The values for which cosø = -1 are ø=n(odd)π, so when\n4πt=n(odd)*π, t = 1/4*n(odd); thus for t = 1/4, 3/4, 5/4, 7/4, 9/4, 11/4 the velocity is zero\n\nThe maximum velocity occurs when cosø = 1, so the max velocity is 8π
786994
velocity-height and horizontal movement. 11.2 km/s in order to escape earth's atmosphere (air provides resistance against movement) above 50 km and movement in horizontal direction so that the object falls perpetually towards the earths gravity centre.
943563
I have no clue of any, but I know that the escape velocity of Earth (the speed of which it takes to escape Earths gravity) is 25,000 mph. Try launching six and a half tons 25,000mph!=)
212948
“the maximum flow rate” means maximum volume of blood flowing per second. \n\nBlood vessel is a cylindrical tube.\n\nVolume of cylinder is area of the circular cross section of the tube x length. \n\nThe length is the distance covered by blood in one second.\n\nIt is given that it is 62 cm/s.\n\nArea of circle is 3.14 x radius x radius.\n ........................= 3.14x0.7x0.7=1.5386 (cm) x (cm)\n\nVolume of cylinder is 1.5386(cm) x (cm) x62 cm/s = 95.4cm^3\n\nThis volume flowing per second is the maximum flow rate  = 95.4cm^3/s.
683205
About 1000 feet per sec.
224790
Anything over 30MPH the speed limit is considered wreckless driving. So if you're going to speed, no more than 29 over!
172859
The terminal velocity of a falling body occurs during free fall when a falling body experiences zero acceleration. This is because of the retarding force known as air resistance. Air resistance exists because air molecules collide into a falling body creating an upward force opposite gravity. This upward force will eventually balance the falling body's weight. It will continue to fall at constant velocity known as the terminal velocity.\n\nThe magnitude of terminal velocity depends on the weight of the falling body. For a heavy object, the terminal velocity is generally greater than a light object. This is because air resistance is proportional to the falling body's velocity squared. For an object to experience terminal velocity, air resistance must balance weight. An example that shows this phenomenon was the classic illustration of a rock and a feather being dropped simultaneously. In a vacuum with zero air resistance, these two objects will experience the same acceleration. But on the earth this is not true. Air resistance will equal weight more quickly for the feather than it would for the rock. Thus the rock would accelerate longer and experience a terminal velocity greater than the feather.\n\nAnother factor that affects terminal velocity is the orientation at which a body falls. If an object falls with a larger surface area perpendicular to the direction of motion it will experience a greater force and a smaller terminal velocity. On the other hand, if the object fell with a smaller surface area perpendicular to the direction of motion, it will experience a smaller force and a greater terminal velocity.\n\nThe terminal velocity for a skydiver was found to be in a range from 53 m/s to 76 m/s. Four out of five sources stated a value between 53 m/s and 56 m/s. Principles of Physics stated a value of 76 m/s. This value differed significantly from the others. Then again, the value is variable since the weight and the orientation of the falling body play significant roles in determining terminal velocity.\n\nAn object which is falling through the atmosphere is subjected to two external forces. One force is the gravitational force, expressed as the weight of the object. The other force is the air resistance, or drag of the object. The motion of any object can be described by Newton's second law of motion, force F equals mass m times acceleration a: \n\nF = m * a \n\nwhich can be solved for the acceleration of the object in terms of the net external force and the mass of the object: \n\na = F / m \n\nWeight and drag are forces which are vector quantities. The net external force F is then equal to the difference of the weight W and the drag D \n\nF = W - D \n\nThe acceleration of a falling object then becomes: \n\na = (W - D) / m \n\nThe drag force depends on the square of the velocity. So as the body accelerates its velocity and the drag increase. It quickly reaches a point where the drag is exactly equal to the weight. When drag is equal to weight, there is no net external force on the object and the object falls at a constant velocity as described by Newton's first law of motion. The constant velocity is called the terminal velocity .\n\nWe can determine the value of the terminal velocity by doing a little algebra and using the drag equation. Drag depends on a drag coefficient, Cd the air density, r the square of the velocity V and some reference area A of the object: \n\nD = Cd * r * V ^2 * A / 2 \n\nAt terminal velocity, D = W. Solving for the velocity, we obtain the equation \n\nV = sqrt ( (2 * W) / (Cd * r * A) ) \n\nwhere sqrt denotes the square root function. Typical values of the drag coefficient are given on a separate slide.\n\nThe terminal velocity equation tells us that an object with a large cross-sectional area or a high drag coefficient falls slower than an object with a small area or low drag coefficient. A large flat plate falls slower than a small ball with the same weight. If we have two obje
778457
75mph! or as fast as you want long as you don't get caught!
1174553
You misinterpreted something. By your theory, when I jump up from the ground, my final velocity is equal to Earth's escape velocity, which is around 9 meters per second, I think. Probably more. But either way, that is much faster than I can possibly go.\n\nAlso, there is NO law which says that velocities higher than c are impossible, it just says that velocities EQUAL to c are impossible.
789497
Between 30 and 40 mph.
743106
a) it is always same speed.\nb) Vf=at+Vi\nVf= -8m/s (because it is going downward)\nvi= 8m8s\na= -9.8m/s^2\n\n-8m/s=(-9.8m/s^2)t+8m/s\n-16m/s=(-9.8m/s^2)t\nt=1.63 s
1111523
I think that "D" is a combination of the drag coefficient, the denisty of air, and the cross-sectional area of the skydiver.  I make that assumtion because they don't provide any of that information, and the units work out.  I don't know if it encapsulates the "2" in the terminal velocity formula or not.  The numbers look prettier if you assume it does.\n\nSo use the formula: Vt^2 = (m x g) / D\n\nVt = Terminal Velocity\nm = mass\ng = acceleration of gravity (9.8m/s^2)\nD = "the constant"\n\nsolve for D:  \nD = (85 x 9.8)/(42^2)\nD = .472\n\n\nfor part 2, solve for Vt:  \nVt = sqrt((40 x 9.8)/.0.25)  \nVt = 39.6\n\nBy the way, the first answer is really really wrong.