While I am falling and moving to the ground there is an acceleration of \n9.8 meter per second every second of  travel time.
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Depends in relation to what. If it's the earth, you could be at rest for a few instants of time before your muscles made you move, maybe to walk or lie down or sit or whatever. However in relation to the sun which is one of the focii of the ellipetical orbit, the earth with us on it is moving along that orbit and takes it a little more than 365 days to complete one revolution. The earth is also rotating around a tilted axis, about 40000 km every 24 hours. The Sun by itself is also moving in a 225 million year orbit around the center of the Milky Way Galaxy of which it is a part, as could the Milky Way around a as yet undetermined obit and period around may be a larger Galaxy or a cluster of Galaxies, which themselves could be moving in another orbit or may be yo-yo back to the big crunch - yes you could be moving even if you were lying supine and still on earth.\n\nI hope that answers your question.
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All objects are pulled by earth with constant acceleration\n\nAcceleration is change in velocity per unit time = change in velocity / time\n\nIn one second the velocity changes by g unit.\nIn t seconds it velocity reduces by gt.\nIf u is the initial velocity and if in t seconds it is reduced to zero then u = gt.\n\nOr t = u/g.\n\nThe time to go up is u/g.\n\nThe height it has gone up is given by average speed x time. = u/2 x t = u/2 x u/g = uu/ 2g.\n\nWhen it falls down it starts with zero speed, but falls with the same acceleration g.\n\nWhen it reaches the ground let its speed is v and time of fall is t2.\n\nv= gt2.\n\nSince the distance is = average velocity x time = v/2 x t2 = gt2/2 x t2= gt2t2/2  \n\ngt1t2/2 = uu/ 2g. or t2t2 = uu/ gg  ot t2 = u/g .\n\nThus time of fall is equal to time of up.
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gravity of 9.8 meters/ sec\n\n\nDistance = 1/2 x Gravity x Time(squared)\n\n43=.5 x 9.8 x t^2\n\nt = 2.96\n\nso \n\n2.16^2 x 9.8 x.5 = 22.86\n\n43-22.86=20.14m\n\n\nI think he is 20.14 m above the ground assuming he does not achieve terminal velocity or hit another branch on the way down.
500194
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
464614
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
260170
Yes, velocity is changing it's direction as well as magnitude. Acceleration due to gravity is acting downward.
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k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
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Velocity (or speed) defines the time it takes to cover a distance from point a to point b.\n\nV=d/t  >\nV (velocity)\nd (distance in meters)\nt (time in segonds)\nm/s (meters per segonds)\n\nAcceleration is a variation in the velocity, wether it increases OR decreases.\n\nie: gravity on earth has an acceleration of \n-9.98m/s  (notice the minus)  so if an object is falling down on earth it will accelerate towards the ground because the object is subject to that force of gravity, \nif it is thrown up it will slow down as it goes up because the same force of gravity affects the speed of the object, when it has slowed down completely it has reached 0m/s and will start to fall again and accelerate untill it reaches a surface that will stop it from falling further (ground, floor, your hand, etc)\n\nThe formula for acceleration is:\n\na= V1-V2/t \n\nor acceleration = initial velocity-final velocity divided by time.\nif it had an initial velocity that is different from the final velocity there was an acceleration, (or change in velocity).
208039
Actaully I am still in flight.
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Since I was standing upon the earth’s surface before I started to levitate, my body would have a certain amount of kinetic energy prior to the levitation.  Upon levitating, this kinetic energy would “propel” me in the direction of the earth’s rotation.\n\nSo would I come down in the same place?  That would depend upon the resistance of my body to the friction of the atmosphere and the direction of the prevailing wind during the time period of levitation.  It might be possible for a very strong wind to push me faster than the earth is rotating or to push me perpendicular to the direction of rotation.  Or even "backwards."\n\nIn an “ideal” experiment, with no atmospheric friction, I should land exactly where I left the ground, give or take the distance of one step.
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For maybe a half a second. The ball will initially go as fast as the car, but, because the mass of the ball has so less mass in comparison to the car, it will quickly lose momentum and slow down and fall to the ground.
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Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
214757
An object in motion remains in motion unless it is acted upon by an outside force.  Say you're traveling in a train going 200 miles an hour, jumping on a trampoline.  Before you jump, both you and the train are moving 200 miles an hour forward.  When you jump, you'll begin moving upward at about 5 mph, but you'll still be going 200 mph forward as well, just like the train.  When you come back down, you'll meet the trampoline exactly where you left, because you, the train, and the trampoline have all been moving forward at the exact same speed.  Here's another example:  The circumference of the earth is around 24,000 miles, and the earth completes one revolution per day.  That means that, in one day, you travel in a 24,000 mile circle, yet you don't feel that you are moving because everything else is moving at the exact same speed.
185348
You may be confusing acceleration with velocity. An object in motion will remain in that motion unless a non-zero net force is applied to it. If you are in the void of space, and are traveling along at a particular velocity, you will remain at that velocity. On Earth, drag and other forms of friction will slow you down.\n\nAcceleration is different. For something to continue to accelerate, a constant non-zero net force must be applied to it.\n\nacceleration = force/mass so if I have a mass of 50kg and want to accelerate at 100m/s^2 (100 meters per second per second) then I need a constant force of 5000 newtons applied to me. Keep in mind that that is 5000 newtons relative to ME.\n\nNow, according to our current understanding of light, it would take an infinite amount of energy to reach speed of light because force = mass/acceleration and as you accelerate, your mass increases! Neither you nor I will notice this effect in our daily life but it becomes more noticable as you approach the speed of light. This does not create mass in the sense that new mass is appearing in the universe, it just adds more weight (no pun intended) to the mass that's already there. This is actully called the reletivistic mass. Your mass when you're at rest, is your rest mass.\n\nThough we won't be traveling close to the speed of light any time soon, there have been plans to get humans to high speeds without killing them. One such way is to accelerate in intervals. For example, you feel a strong force when taking off in an airplane but once the airplane reaches a few hundred miles per hour, and you've caught up, you can do it again.
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I would have time to land on my feet. I would just feel an adrenaline surge!
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Here's your answer:\n\nForce = mass * acceleration\n\nacceleration = change in velocity / change in time\n\nWe want to know this change in time, so substitute the second equation into the first:\n\nForce = mass * (change in velocity / change in time)\n\nWe know:\n\nForce = -3000 N (negative because it acts to decrease our speed)\nmass = 1000 kg\nchange in velocity = -6 m/s (we are going from 6 m/s to 0 m/s)\n\nSolving for the change in time, we have:\n\nchange in time = (1000 kg) * (-6 m/s) / (-3000 N)\n\nTherefore, the time = 2 seconds
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2. An oxygen molecule moving at 560 meters/second.