an object under the vaccum condition in heigh direction has thrown up ward, after 4s came back to the first place. indicate how heigh this object went up . and with witch speed this object has thrown? (answer 19/6 m/s.  19/6 m)
200740
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
350371
kaksi_guy: I think charlesellis753 is correct.  If we interpret the question as a fill in the blank, mass appears to be the best answer.\n\nIf the ______ of an object doubles, then the force required to accelerate it at the same rate will double.\n\nThat sentence is true is the blank is filled with 'mass'.
122992
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
493187
1st step: OK we need to know how many times  faster the car went on the way to the service station than on the return trip:\n49/16=3.0625 times faster on the way up than the way back. \n\n2nd step: find the how much time it took on the way up and the way back. \n\nNow comes the Algebra:\nif we let X stand for the way up, the equation for the round trip would be:\n X+ 3.0625X= 4\n4.0625X=4\nX=4/4.0625\nX=.985 \n\nso the time up = .985 hours and the time back = 4 -.985, or 3.015 hours.\n\nso how far is the station? well, it's 49km/h x .985 h, (using the data for the trip to the gas station) or:\n\n48.265 km\n\nok we're DONE!!\n\nPS: I have a sneaking suspicion that if you look at the problem again, you might find it said 48 km/h on the way up, which would work out to be that the trip was EXACTLY 3 times faster on the way up. ok just a hunch. But from what you gave me, that is the answer.
179473
yes, if u throw any object away from u , u will be propelled in the opp direction by a velocity, v= mu / ( M + m ) where u is the velocity of the thrown object, m is mass of object and M is ur mass.\n\nthis is conservation of linear momentum for u!!
279221
C, gravity is an acceleration, it is by definition not uniform then.  \n\nit's 9.8 meters per second per second, just in case you wanted to know, that's also written 9.8 m/s^2\n\nhere's a little more: it's not A, (see above), it's not B, since again this would mean that it's uniform, since 10 seconds is twice as great as 5 seconds, and it's not D because it would fall faster in a vacuum than in air since there would be no resistance.  if you drop a pen and a piece  of paper right now, the pen should fall faster since the air will resist the paper more, but in a vacuum both would hit the ground at the same time.
218060
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
75670
look at the question and it provides the answer --- it is acceleration occurring at a uniform rate (ie. it is constant).  for example, from a stationary start, a uniform acceleration of 2m/s^2 will give speeds of 0s=0m/s, 1s=2m/s, 2s=4m/s, 3s=6m/s.
384201
To figure out how long it will take the object to hit the ground, we can use the distance equation: d=1/2at^2 ---> t=√2d/a\nThe distance is 100m, and the acceleration is 9.8m/s^2: \nt=√2(100)/9.8 --->t=4.5s\nNow that we have the time it's in the air, we can figure out the horizontal distance by using the velocity equation: \nv=d/t --->d=vt\nThe velocity is 20m/s, and the time is 4.5s, so: \nd=(20)(4.5) ----> d=90m\n\nAs for question C, the objects horizontal velocity (ignoring air resistance) will stay the same througout its flight, so it hits the ground at the same horizontal speed it had when it started: 20m/s\n\nAnd to figure out question D, just remember that velocity will equal acceleration times time. So: \nv=at ----> v=(9.8)(4.5) --->v=44.1m/s\n\nHope this helps!
511051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.
5497
(a) it's an inelastic collision.  An inelastic collision is when they stick together, and an elastic collision is when they bounce off of each other\n(b) p=mv  momentum= mass x velocity.  p= 30 m/s x 500kg = 15,000 kg*m/s\n(c) well, momentum is conserved, so the we would use the formula from before and the momentum(p) from before, except that now the mass is different because it is the combined masses of the two vehicles (300+500=800) , so you can solve for the new velocity.\np=mv  15000 = 800 x v\n\ndivide both sides by 800 to get  v=15000/800 = 18.57 m/s
412010
This is really a calc question.\n\nAn item will fall at 32ft per sec per sec from a rest.\nh - the free fall = t\n\n2. As it is stated the axle is frictionless, therefore the cyinder and axle will move and there is no force between the two. However there is a force between the rope and or bucket and the cylinder.\n\ndouble check the 32ft/s in your book, there may be more to the equation.\n\nCarl
82392
It's around 23.6 mph (penny falling flat) to 73.3 mph (penny falling on its side).\n\nVterm =  sqrt ((2 * m * g) / (c * p * A))\n \nwhere:\nm=mass of the object \ng=acceleration from gravity \nc=coefficient of drag \np=density of air \nA=Area of the object \n\nIt's pretty straightforward.  The full explanation is at the source.
461705
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
449637
the correct answer is not listed ( 32 would be correct if box 2 did not move ) \n\nsince the friction of the surface is not given the correct answer is unknown - it would be the amount of force required to overcome the friction OF THE SECOND BOX ONLY ( you said BETWEEN the boxes and when in motion force is proportional to resistance )\n\nor you did not phrase the ? properly ( likely )
198058
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
384322
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s