A steel ball whose mass is 100g is rolling at a rate of 2.8m/sec. What is its momentum?
473588
Momentum is mass times the velocity.\n\nMass is 100g = 0.1 kg\n\nvelocity is 2.8 m/s.\n\nMomentum is 0.1 kg x  2.8 m/s. = 0.28 kg.m/s
253742
Please take the time to\nvote for the best answer\n\nThank You!!\n\nRecall that an object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object: \n\nLook this up for more
770662
Let m be the mass of the block and M be the mass of the ball.\n\nThe force pulling the block downward mg sin 35 = 60g sin 35 = 34.41 g\n\nThe frictional force opposing this force is 0.22 x 60 g cos 35 = 10.81 g\n\nThe weight of the ball is also opposing this force = Mg\n\nSince the block is moving with constant speed the net force is zero,\n\n34.41 g - 10.81 g - Mg = 0\n \n M = 23.6 kg.
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
824387
The ball goes from 0 to 9/s in .15 secs\n\nSo:    9meters           1\n        ------------  x    ------                  =9 /.15 m/sec2\n           second          .15secs\n\n                                                      =60 m/sec 2
730564
1. friction\n2. W=mg  , he has a mass of 100Kg. Weight is the force=newtons, g=pull of gravity N=100 x 9.8m/sec^\n98Newtons\n3. correct
833966
The question is easy enough if you think about it properly.  What you want is the speed at which the centrifugal force (given by mv^2/r) equals the gravitational force given by the bank, and that will be mg sin(2.5 degrees).  Note that m cancels out in this, and you are given r, so v is the only unknown and you can solve for it with a scientific calculator.
1035224
v^2 = 2gh\n\nh=1/2 gt^2\n\nsubs this.\n v^2 = 2g (1/2 gt^2 )\nv^2 = 2 g^2 6^2\nv^2 =  100 X 36\n\nv= 60m/s
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
938540
assume shot putter moved the shotput with constant acceleration\nhence we need to find acceration\n\nv^2= u^2 + 2as\n(13)^2= 0 + 2a (2.8)\n169 = 5.6a\na= 30.17857142m/s^2\n\naverage force exerted \n= ma\n= 7 x 30.17857142\n= 211.2N\n\nyey
888670
First, draw a picture. Let C be Chris, B be Blaire, and A be the ball at some point. Thats a right angled triangle, where side BA is how far the ball has dropped; CB = 100 feet, and CA = the fishing line.\nNow, we need to know how fast the ball is dropping. Since its distance is 16t^2, its velocity is 32t (differentiating). So side BA is changing at 32t feet per second at any time.\nWe want to know the rate at which side CA is changing. But we know that CA^2 = CB^2 + AB^2, so CA = sqrt(CB^2 + AB^2) = sqrt(100^2 + AB^2).\nNow, we need to differentiate both sides with respect to time.\nThus, d/dt(CA) = d/dt(sqrt(100^2 + AB^2)).\nWe need the chain rule for the RHS: sqrt(something) differentiates to 0.5(something)^-0.5, then we multiply by the inside differentiated. AB^2 differentiated (with respect to t) is 2*AB *d/dt (AB) = 2*AB*32t.  So the overall equation becomes:\nd/dt(CA) = 0.5(100^2 + AB^2)^-0.5 * 2 * AB * 32t.\nNow, we know t = 3 and AB = 16*3^2, so substitute that in and you'll get the answer of 78.9 feet/second.\n\nI'm likely to have made a mistake, but hopefully that gives you the idea of what to do.
433903
The ball will reach a maximum height, then the velocity of gravity (9.8 meters/second) would cause it to travel downward until another force acts upon it (according to Newton's laws of motion). However, this does not account for other forces, such as air resistance. The rate at which the ball falls and the air resistance acting upon it will depend on the properties of the ball (surface area, weight, etc.). If the ball is inside a vacuum, it is said to be in free fall (as the only force acting upon it is gravity) and there is no other force acting upon it. If the ball is allowed to fall long enough, it will reach a point of terminal velocity, where it cannot fall any faster (the force of air resistance will equal the weight of the object)
756774
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
655182
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.
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.