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A spaceship leaves Earth and maintains a constant force By means of a nuclear engine. As the speed of the spaceship increases, an observer on Earth finds that relative to her the magnitude of the spaceship's acceleration is a. 0. b. decreasing constant. increasing proportional to the kinetic energy of the spaceship.

Respuesta :

Answer:

decreasing

Explanation:

This can be explained based on Einstein's theory of relativity and the relation is given by:

[tex]M = \frac{m}{\sqrt1 - (\frac{v}{c})^{2}}              (1)

where,

M = relative mass of the spaceship

v = speed of the spaceship

c = speed of light in vacuum

m = rest mass

Also from newton's second law we know that:

F = Ma

Thus

a = [tex]\frac{F}{M}[/tex]                  (2)

Using eqn (1) and (2):

[tex]a = F\frac{\sqrt{}1 - (\frac{v}{c})^{2}{m}}             (3)

From eqn (3):

We can observe that on increasing the acceleration of the spaceship its velocity decreases.

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