Imagine you tied a rope around the earth's equator, and that you cut it and added one metre to its length. If you could somehow suspend the rope above the ground so that it floated like a ring around the planet, HOW BIG WOULD THE ROPE BE BETWEEN THE ROPE AND THE GROUND?
217711
It would be 1 meter divided by 2pi, or 0.159 m (159 mm) approximately.
243900
It's about 1/6 earth gravity. (But don't forget to add in 1/6 of the spacesuit you have to wear, too.)
207313
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
76407
The distance around it.The circumference.
587823
Has anyone notices that the answer given by "d_1167" has absolutely nothing to do with the question which was asked?\nIn fact, his answer was not even his own, he merely copy-pasted it from a web site and then did not even have the decency to site it,\nhttp://education.vsnl.com/karnik/velocity_of_light.pdf\nor the HTML version,\nhttp://66.218.69.11/search/cache?p=education.vsnl.com%2Fkarnik%2Fvelocity_of_light.pdf&ei=UTF-8&fr=FP-tab-web-t&x=wrt&u=education.vsnl.com/karnik/velocity_of_light.pdf&d=FNtzZDmtMwG4&icp=1&.intl=us\n\n\nAnyway, to get back to the question at hand,\nI would have to lean somewhat towards the direction of the answer given by "NotEasilyFooled".\nThe speed at which the effects of cutting the rope on planet M would not reach the 2nd planet (N) until at least 1 year after it happened since no information can travel fast than light.\n\nSince the rope is massless (but we assume it to act as a normal rope anyway), there is no reason to think that it would necesarily fall the the surface of of planet N except for the fac that there was a force of tension in the rope when it was cut, pulling it towards the surface of planet N.  But we already assumed it was massless (or at least "weigthless")....applying a force to a massless object puts us in a rather awkward position (F = ma).
531463
The net force acting on the bucket by Newton's Second law =\n3.9X2.8 = 10.92 N upwards.  Forces acting on the bucket are 3.9x9.8 = 38.22 N downwards by the earth and F, the force acted upon by the rope on the bucket which is upwards. So we have\n\nF - 38.22 = 10.92 or F = 38.22+10.92 =  49.14 N is the final answer.
86424
They want you to use the pi = C/d formula.  You have two circles, one whose circumference (C) is 1 yard longer.  You need the earth's circumference in yards.  You know pi, you know C, so calculate d for both the earth and for the earth + 1 yard.  The difference (very tiny) is the distance the string will be from the earth's surface.
15320
The force on the system is 3.0 kg x g. (the only force is due to gravity acting on the hanging mass) The mass of the system is 3.0 kg + 5.0 kg = 8.0 kg.  The acceleration of the system is a = force/mass = 3.0 kg x g / 8.0 kg   =  3 / 8 g or 0.375 x 9.8 = 3.675 m/sec^2
674585
No little shaver. You do not have enough guts to circle the globe. The earth is (I am guessing) only about 30,000 miles in diameter. I think (but do not know) that you have about 30 ft of intestines. Certainly not as much as even 30 yards. How do you suppose the earths population of billions would even have teh room for each of us to carry a globe-belt in our guts.What do you think it is made of?Dental Floss? You might (might) have enough nerve paths to circumvate the globe. but I doubt it. They might have meant blood vessels including the capillaries. But I doubt that one too.
322767
Already answered "3" privately, but here it is for others:\n\nAssumptions:  Weight of rope (or cable) is neglected\n              Both ends rigidly fixed\n              Rope acts elastically with a stiffness of EA\n\n              To = Initial tension\n              T1 = Deflected tension\n               a = Angle rope makes with initial position \n               W = Force applied to middle of rope.\n\nTwo equations to satisfy:\n\n1) 2T1*sin(a) = W\n\n2) T1 = To + (sec(a) - 1)*EA\n\nEliminating T1:   To + (sec(a) - 1)*EA - W/(2sin(a)) = 0\n\nSolving for "a" (I had to do it numerically!) and plugging that value back into eq 1) will give the final tension.  The length does not affect the answer, only the initial tension and stiffness of the rope.\nAmazing
14748
well whatever force that is being exerted on the rope, that is going to be the tension. that means that all you have to do is calculate the force.  you just have to realize that not only is the gymnist accelerating, producing force, she is also overcoming the force of gravity. therefore, if you were to draw your free body diagram, you would have 2 forces pointing down, the force of gravity(mass times gravity), and this acceleration force(which would just be mass times the acceleration.  if you add these two forces together, you get the total force being exerted on the rope which is equal to the tension. here is the equation:\n\nT=F + Fg  (where T is tension, F is the force caused by the acceleration, and Fg is the force due to grabity.)\nT= ma + mg\nT= (74)(.11) + (74)(9.8)\nT= 733.3N.\n\nthis can also be done by just adding the acceleration to the acceleration due to gravity then multiplying this new acceleration by the gymnist's mass. either way you will get T=733.3N
43767
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
47845
It could be the grid of the earth, (latitude and longitude) the equator is the x-axis. and so on
477030
None of the above....\nSaturn's rings are (presumably) in orbit around that celestial body.  As such, they (presumably) would maintain an equal distance from the (centre) of the planet.  They would, by definition, be always 'widest open' as you say above.
404563
Consider a mass m on the surface of \nEarth. The equation governing this mass\ncan be written as\nmg=G*m*M/R^2\nm=mass of the object=kg\ng=acceleration due to gravity=9.81m/sec^2\nM=mass of Earth kg ?\nR=radius of Earth=6.4*10^6m[given]\nG=universal constant of Gravitation\n=[6.6742+-.0010]*10^-11Nm^2kg^-2\nCanceling m we can write\nM=g*R^2/G\nYou have all the values\nSubstitute and get Mass of Earth.
461303
since it is a smooth vertical circle i.e.\npie 22/7,the let the highest point of the string to the lowest be h,given: ring=w,angle=2a\ntherefore the ring will go wobbly down and the string will tighten.
437644
how long is  apeice of string???\nwouldnt it be good to see what is lurking at the bottom of the sea. could be a whole new world!! (well you never know!)