Pregunta
A very small round ball is located near a large solid sphere of uniform
density. The force that the large sphere exerts on the ball
a. can only be calculated using calculus.
b. is independent of the mass of the ball.
c. is approximately the same as it would be if all thi mass of the sphere
were concentrated at the center of the sphere.
d. is exactly the same as it would be if all the mass of the sphere were
concentrated at the center of the sphere.
density. The force that the large sphere exerts on the ball
a. can only be calculated using calculus.
b. is independent of the mass of the ball.
c. is approximately the same as it would be if all thi mass of the sphere
were concentrated at the center of the sphere.
d. is exactly the same as it would be if all the mass of the sphere were
concentrated at the center of the sphere.
Ask by Dickson Edwards. in South Africa
Nov 26,2024
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
a. The force can be calculated using calculus.
b. The force is independent of the ball’s mass.
c. The force is approximately the same as if the mass is at the center.
d. The force is exactly the same as if the mass is at the center.
b. The force is independent of the ball’s mass.
c. The force is approximately the same as if the mass is at the center.
d. The force is exactly the same as if the mass is at the center.
Solución
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Como
Beyond the Answer
The gravitational force exerted by a uniform sphere on an object outside it behaves in a fascinating way: it acts as if all of the mass were concentrated at its center. This principle is a consequence of the shell theorem, which states that the gravitational forces from different parts of a spherical mass cancel each other out for points outside the sphere. So, for the small ball, the force can be calculated more simply than it might seem!
In practice, this knowledge allows engineers and physicists to simplify complex situations. For instance, when designing satellite orbits around planets, they can treat the planet as a point mass at its center. This understanding not only streamlines calculations but also helps ensure that space missions are efficient and accurate, keeping everything running smoothly in the cosmos!
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