Question

A particle is given an initial speed u inside a smooth spherical shell of radius

Answer 1

Explanation:

The speed will remain constant as the sphere is smooth and no friction will occur with the surface of the particle and the surface of the sphere....

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Answer 2

$\huge\bf{Answer:-}$

Let u be the speed.

Let r be the radius.

$u {}^{2} = 5gr \\ v {}^{2} = u {}^{2} - 2gr \\ = 5gr - 2gr \\ = 3gr \\ = a(t) = g \\ = a(c) = v = v \frac{2}{r}$

$a(c) = \frac{v {}^{r} }{r} = \frac{3g {}^{r} }{r} = 3g$

$Acceleration = \sqrt{a(t) {}^{4} } + a {}^{2} (c)$

$= \sqrt{g {}^{2}+ 3g {}^{2} }$

$= \sqrt{g {}^{2} + 3g {}^{2} }$

$= g \sqrt{10}$