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mass = weight ÷ gravity Since the gravitational pull is relatively constant near the surface of the earth, you can weigh the object, then divide the weight by the gravitational acceleration (9.8 m/sec2 near the earth's surface).
First you have to convert weight into mass. This is dependent on the acceleration the mass is experiencing (either gravitational or centrifugal). If it is gravitational and it is at or near the surface of the Earth then mass=weight/9.81m/s2 If it is centrifugal then a=v2/r and mass=weight*r/v2 Then to find momentum just multiply mass by velocity.
To calculate this speed, you need some more numbers, not just the distance from Earth to Sun. You need to know:* The gravitational constant * The distance from Earth to Sun * The mass of the Sun You DON'T need the mass of the Earth. Assume any mass; you can just call the mass of the Earth "m". Then calculate an expression for the gravitational attraction between Earth and Sun. Divide by the mass of the Earth, and you get a centripetal acceleration. Assuming a circular orbit for simplicity, the centripetal acceleration must be just this force. Use the formula for the centripetal acceleration along a circle (a = v squared / radius). (Don't forget to convert the distance from Earth to Sun, to meters!) Solve for "v".
It is equal to 308.64 pounds approximately. Kilogram is the metric unit and pound is the imperial unit for mass. 1 Kilogram is 2.204 pounds. So we multiply kg by 2.204 to get the equivalent pounds.
If you are ignoring energy lost due to friction, the total mechanical energy will be the same after it has traveled 1 meter as when it was dropped. This means the easiest way to solve the problem is to find the mechanical energy at the beginning, when the ball is at rest and all of its mechanical energy is gravitational potential energy. Gravitational potential energy equals mass*g*height. Since mass*g equals weight, we can just multiply 10N by 4m, making the total mechanical energy 40J.After it has traveled 1 meter, some of the gravitational potential energy has been converted into kinetic energy. The gravitational potential energy is just the weight of 10N multiplied by the height of 3m, or 30J. To find the kinetic energy, we need to find velocity2, which equals 2 times acceleration (g) times displacement (1m) when the initial velocity is 0. We also need the mass, which is weight (10N) divided by g. Kinetic energy equals (1/2)*mass*velocity2, so we get (1/2)*10N÷g*2*g*1m, which equals 10J, so the total mechanical energy is still 40J.
You need to know the mass of the solid.
You need to use the radius and the mass :P
You need to know . . . -- the mass of each object -- the distance between their centers of mass -- the value of the universal gravitational constant
Gravitational energy is U = mgh. Therefore, you need the egg's mass and the height at which it will be dropped. g=9.8
No. The gravitational force is a different force from magnetism, and depends only on the mass and the distance. Specifically, a body does not need to rotate to have gravitational force.
You need to have a weight and the mass of an object then you use the formula f=w=mg
The book has a mass of 0.46kg
To calculate the gravitational PE, you also need the height. PE = mgh (mass x gravity x height); convert the mass to kg. (3 mg = 0.000 003 kg); use 9.82 for gravity; and convert the height to meters. The result will be in Joule.
You need to compress a large mass into a very small space. As far as we know, this can only be achieved by the gravitational collaps of a star.
Look at the formula for gravitational potential energy: PE = mgh. It is the product of mass, gravity, and difference in height. Therefore, those three factors are required.
The gravitational field strength at a standard distance is directly proportional to a planet's mass so the need for a scatter diagram is not immediately obvious.
Everything that has mass exerts a gravitational pull on everything else with mass... but gravity is such a weak force that objects need to be huge in order to have any noticable effect. No person will ever be big enough that they can noticably attract objects around them.