Aurora wrote / s: Greetings! We could use help in solving the following tasks: 10 m long railway wa

I have a problem in solving these tasks in which the amount of circulating satellite with a mass of 2.5 tonnes, which encircles the earth in 143min and 20s? The radius of the earth is 6400, the gravitational constant 6.67 * 10 per -11Nm / kg per square mass of the Earth is 6 * 10 24 Looking at the height of the satellite ... teja33 Posts: 2 Joined: 18/09/2007 16:09
That we do not all write again, I enclose a link that will help you: http://forum.kvarkadabra.net/viewtopic.php?t=1616 The first task is the same as yours, except that instead of seeking a radius of gyration of patrol time. Zdravapamet Posts: 2840 Joined: 08/16/2004 18:41
That is to say. We know the mass of the satellite and the satellite bypass time. The radius of the Earth to be, the gravitational constant and the mass of the Earth. Satellite to circulate away from the surface of the earth, that is the radius of the track and that we were looking for. On the satellite operates gravitational force (the Earth is equal to the force acting only in the other direction) according to Newton's gravitational law: On the other hand, from the mechanics of rotation you know that the force that the body awarding the necessary acceleration for uniform circular motion, size: When calling a word and equation množiš with, you get: Upoštevši immediately follows bromo or sought after height: zdravapamet Posts: 2840 Joined: 16/08/2004 18:41
Greetings! I need help with the following tasks: What is the bypass time satellite bromo that orbits the Earth at low altitude? Note: The centrifugal force equal to the force of gravity, the satellite! Thank you for any help. Bojan 0bojan0 Posts: 2 Joined: 01/14/2008 16:03
Greetings! We could use help in solving the following tasks: 10 m long railway wagon is moving steadily at a speed of 20 m / s. The bridge, which is 10 m above the rail, the horizontal stone is thrown at the time when the front part of the wagon runs under the bridge. What should be the maximum and minimum extent to which the initial velocity of the stone, the stone falls on the wagon? Solutions: v1 = 13 m / s v2 = 20 m / s Thanks already in advance to help with the task. Aurora Posts: 2 Joined: 12/03/2007 19:21
Aurora wrote / s: Greetings! We could use help in solving the following tasks: 10 m long railway wagon is moving steadily at a speed of 20 m / s. The bridge, which is 10 m above the rail, the horizontal stone is thrown at the time when the front part of the wagon runs under the bridge. What should be the maximum and minimum extent to which the initial velocity of the stone, the stone falls on the wagon? Solutions: v1 = 13 m / s v2 = 20 m / s Thanks already in advance to help with the task. When the stone is a horizontal throw, the wagon by uniform motion. These are two examples: 1. A stone is dropped to the rear of the wagon. 2. Rock fall on the front part of the wagon. In the second case, the same range of stone paths made by the wagon:. In the first case, the range of the stone for the wagon length () is shorter than the path made by the wagon: When inserted, we get:. shrink Posts: 9735 Joined: 4/9/2004 17:45
Please help with one task: at the equator of a spherical planet weighs less than twice the body to the pole. The density of the planet is 3x10na3 kg / m 3. Calculate the time in which the planet is subjected to a single revolution around the axis. The solution is: 2 h 41.6 min. Anya Posts: 166 Joined: 13.05.2009 15:14
On the pole weighs at the equator, however: Note that the resultant of all forces acting on the body (as much as shown by the scale), mass times the radial acceleration of the body: From this we get: Masi body other out, we get: considering that and to give : The term bypass time, if something is not understandable question Naoki Posts: 66 Joined: 12.11.2008 23:02
Anya wrote / s: Please help with one task: at the equator of a spherical planet weighs less than twice the body to the pole. The density of the planet is 3x10na3 kg / m 3. Calculate the time in which the planet is subjected to a single revolution around the axis. The solution is: 2 h 41.6 min. Have you tried yourself? bromo Tip: On the equator due to the centrifugal effect subtracts the weight of the centrifugal force. This difference (based on the data) corresponding to half the weight. Weight over the course write Newton's gravitational bromo Act; the mass of the planet you get to množiš density by volume of a sphere. From here, you get a ratio which is multiplied with a straight bypass time of rotation about the axis. PS I see you've already got a complete solution. shrink Posts: 9735 Joined: 4/9/2004 17:45
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