Physics question below?
827118
We can't use conservation of energy because this is an inelastic collision. Conservation of momentum does hold, so\n\nm1*u1 + m2*u2 = (m1+m2)*v\n\nu2 is zero. So \n\nv = m1*u1/(m1+m2) = .24 kg*u1/2.85 kg = .084*u1\n\nThe projectile's incident kinetic energy \nKEp = (1/2)*m1*u1^2 = 0.12*u1^2 joules\nThe final kinetic energy \nKEf = (1/2)*(m1+m2)*v^2 = 1.42*(.084*u1)^2 = .0100*u1^2 joules.\n\nPercentage = 100%*KEf/KEp
422440
the wavelength is half...\n\nvavelength and frequency are inversely proportional
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
976575
Hi Mark, I will answer your question if you answer mine first.\nWhy are you asking this question, it seems like a lot of people in school are trying to take the easy way out of a physics or math problem.\nWill answer if you are not doing said above!\nCheers, Susie
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
295799
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
648817
a) Supposing that s0 = 0. We  have v0 = 32ft/sec\n=> s = -16t^2 + 32 t\n\nb)When t = 1, s=-16 + 32 = 16 ft above the ground\n\nc)When s = 0 again:\n0 = -16t^2 + 32 t\n=>t(-16t + 32) = 0\nThe equation is equal to zero when t=0 (the time the ball was thrown) and when t = 2 => The ball will hit the ground at t=2\n\nd)ds/dt = -32t + 32\nThis is equal to 0 when t=1\nat this point s = 16.
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
34477
It is 'B'. But I don't understand why you put up this question under 'Physics'!
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
905559
This seems more like a physics question than a chemistry question.  I would try posting this in the physics section- someone will be able to help you there!
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)