A bowling ball (mass = 7.2 kg, radius = 0.10 m) and a billiard ball (mass = 0.49 kg, radius = 0.028 m)?
112606
Whatever sound is made when a brain fries into sludge just came out of my skull.
97851
The ball weighs 14–15 ounces (397–425 g).
930941
12, same as a classical soccer ball. \nIt is a dodecahedron whose edges are capped with equilateral hexagons.
320204
That question makes no sense.  A ball, or sphere, is a 3-dimensional object.  A circle is a 2-dimensional object and doesn't exist in real life.  A ball is not a circle.
998671
Gravity slows down the ball at 9.8m/s/s. After 1s, its velocity is 12-9.8=2.2m/s
438041
Since they r similar,\n\nDivide the Area of the larger cylinder by the smaller one,\n49(3.142)/25(3.142) to receive proportionality\n\n= 1.96\n\nTherefore 1.96 * volume smaller cylinder,\n\n1.96*50(3.142) = 307.8 = 98(3.142) [The Volume of the larger cylinder]
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet
831197
Equation:\n\nV^2 = U^2 + 2 A (delX)\n\nThen solve F = M A\n\nwhere V = Final Velocity\nU = Initial Velocity\nA = Acceleration (unknown)\ndelX = distance of applied force (delX meaning delta X, or change in distance)\nF = Force\nM = Mass\n\nOf course you have to consider two things:\nThe force applied over the two feet would have to be constant (this is an assumption you have to make to solve)\nAll your units are in ft and weight.  you have to use the meters measurements and convert your ball's weight of 0.25 lb to kilograms (that's the mass measurement)\n\nyou now have more than enough information to solve this problem on your own.\n\nNOTE: Jim, that is a horrible misuse of the wrong equations for solving this kind of problem.  Nice try though.
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
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
579002
First I'll assume this isn't a trick question, i.e., that the balls start with the same speed and have equal densities. They are rigid and roll on a rigid surface. What slows them down? Air resistance and going uphill. Not friction because they are rolling, not sliding. \nWith the same initial velocity, they have the same translational momentum per unit mass. However, another factor is the angular momentum due to rolling. Let ball 1 have a radius r=1 and a mass m=1, and ball 2 have r=2 and consequently m=8. If each ball has an initial velocity of 1, ball 1's angular rate w=1 rad/sec and ball 2's w=0.5 rad/sec. The moment of inertia I of a solid sphere = k*m*r^2 (k=0.4), or I=k for ball 1 and I=32*k for ball 2. The angular momentum = I*w, thus I=k for ball 1 and I=16*k for ball 2. While both balls have the same translational momentum per unit mass, ball 2 has twice the angular momentum per unit mass and therefore a greater total momentum per unit mass, which means a greater force * time per unit mass is needed to stop it. Going uphill provides the same force per unit mass on each ball, so ball 2 goes further uphill. Air resistance per unit mass is greater for ball 1 since air resistance depends on area, and ball 1's area/mass ratio is 1 vs. ball 2's ratio of 0.5. Again, ball 2 goes farther.\nEDIT: The balls have different momentum per unit mass but the same energy per unit mass, so my argument above about uphill travel is wrong; they go the same distance uphill. But air resistance still favors the larger ball. So answer 5 is the best answer so far. Another real-life factor would be: suppose the balls are on a rough surface or face an effective drag of a surface such as a rug. A larger ball wins on both. The larger radius means smaller vertical motions for a given scale of roughness (e.g., gravel size), so less energy wasted per unit mass. And you can visualize how a deep-pile rug would stop a rolling golf ball much faster than a bowling ball.
313310
1.  Weigh all 12 balls together.\n2.  Weight 11 balls.\n3.  Remove a ball from the 11 and weigh the 10.\n\nNow for some calculation.\nweights 1,2,3 --> w1, w2, w3\n\nif w1-w2 = w2-w3 --> you have the weight of a normal ball.  To find the weight of the odd ball, multiply the normal weight by 11 and subtract from the weight of 12 balls.\n\nif w1-w2 does not equal w2-w3 -->  one of the two differences is the weight of the odd ball.  You have to assume one of these differences is the weight of a normal ball and analyze.\n  \nTake the larger difference and multiply by 11.  If the difference between this value and the combined weight is equivalent to the one of the differences from the previous paragraph, then the weight of the oddball is the "smaller difference" we havent analyzed yet.  The oddball is thus lighter than the rest.  \n\nIf the subtraction from the previous step isn't equivalent to the smaller difference, then the weight of the oddball is the one you've assumed to be the normal ball.  The oddball is thus heavier than the rest.\n\nOriginal solution, may be kind of complicated, but hey it's orginal, LOL\n\nTo answer below: How would that work if the odd ball is LIGHTER not heavier?
1096423
Look down below, and that should give u a starting point..\nA cue and 2 balls.
1157509
The ball is: \nspherical; \nmade of leather or other suitable material; \nof a circumference of not more than 70 cm (28 ins) and not less\nthan 68 cm (27 ins); \nnot more than 450 g (16 oz) in weight and not less than 410 g\n(14 oz) at the start of the match; \nof a pressure equal to 0.6 – 1.1 atmosphere (600 – 1100 g/cm2) at\nsea level (8.5 lbs/sq in – 15.6 lbs/sq in).
409505
A soccer ball of mass 0.43 kg leaves the foot of the kicker with an initial speed of 25 m/s.The weight of the ball at the start of the game shall not be more than 450 g nor less than 410 g.\nSource-Sam
14955
(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
749375
You mean at the same rate - 32.2 feet per second per second.\n\nGravity pulls on all matter at the same rate. If you drop a bowling ball taped to a baseball, they will fall at the same rate. But it is not because the bowling ball is pulling on the baseball, it is because gravity is pulling on both of them. If the tape breaks gravity will still be pulling on both of them at the same rate. They will hit the ground at the same speed, however, they will hit with different forces. Because Force = mass x acceleration. The acceleration is the same for both of them. But the mass of the bowling ball is larger than the mass of the baseball, so it will hit the ground with more force.