Science, asked by TANU81, 1 year ago

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Answered by ShivajiMaharaj45
2

Explanation:

\sf Question\:14 \\ \\ \sf According\:to\: Rutherford\: experiment \\ \\ \sf \frac {m{v}^{2}}{r} = \frac{kZ{e}^{2}}{{r}^{2}} \\ \\ \sf m{v}^{2} = \frac{kZ{e}^{2}}{r} \\ \\ \sf K.E. = \frac {m{v}^{2}}{2} = \frac {kZ{e}^{2}}{2r} \\ \\ \sf P.E. = - \frac {kZ{e}^{2}}{r} \\ \\ \sf T.E. = P.E. + K.E. \\ \\ \sf T.E. = \frac { kZ{e}^{2}}{2r} - \frac {kZ{e}^{2}}{r} \\ \\ \sf T.E. = -  \frac {kZ{e}^{2}}{2r} \\ \\ \sf so\:the\:Total\:energy\:is\: negative \\ \\ \sf \bold{\underline{\underline{Correct\:option\:is\:B}}} \\ \\ \sf Rutherford \:in\:1911 \:Carried\:out\:an\: \\ \sf experiment\:called\: \alpha-ray\:scattering \\ \\ \sf And \:therefore\:he\:concluded\:that\:mass\:of\:the\:atom\:is\: concentrated\:at\:its\:centre \\ \\ \sf Also\:+ve\:charge\:is\:concentrated\:at\: centre \\ \\ \sf \bold{\underline{\underline{Correct\:option\:is\:C}}}\\ \\ \sf Question\:9\\ \\ \sf Circular\:motion:Circular\: motion \:is\: where\: the\: object\: travels \:in \:a\: path\: which\: is \\ circular\:-can\: be\: a \:full \:circle,\: can \:be \:a\: part\: of\: the\: circle\:\\ \\ \sf Periodic \:Motion:  Periodic \:motion \:is\: where\: the\: object \:repeats\: its \:motion \:after\: a\: particular \:time \\ \\ \sf So \: circular \:motion\:may\:be\: periodic \\ \\ \sf \bold{\underline{\underline{Correct\:option\:is\:A}}}

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Answered by Anonymous
3

Answer:

14:- option(b)- negative.

15:- option(c)- both mass and positive charge are densely concentrated.

9:- option(a)- may be periodic.

Explanation:

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