Question

$\mathtt{\bf{\underline{\red{Need\:full\explaination\:}}}}$

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Answer 1

$\tt Initial\: velocity \:(u) = 0 \: m\:{s}^{-1} \\ \tt Acceleration \: ( a ) = 2\:m\:{s}^{-2} \\ \tt Time\:for\: accelerated\:motion = 10\:s \\ \\ \tt We\:know\:that \\ \\ \tt v = u + at \\ \\ \tt \longmapsto v = 0 + 2 × 10 \\ \\ \tt \longmapsto v = 20 \: m\:{s}^{-1} \\ \\ \boxed{\bold{\underline{\green{\tt The\:maximum\:velocity\:reached\:is\:20\:m\:{s}^{-1}}}}} \\ \\ \tt For\:this\: accelerated\:motion \\ \\ \tt Distance\: travelled\:be\:{s}_{1} \\ \\ \tt {s}_{1} = ut + \frac{1}{2} a{t}^{2} \\ \\ \longmapsto \tt {s}_{1} = 0 + \frac{1}{2} × 2 × 10 \\ \\ \tt \longmapsto {s}_{1} = 20\:m \\ \\ \boxed{\bold{\underline{\red{\tt The\:distance\:travelled\:in\:accelerated\:motion\:is\:10\:m}}}} \\ \\ \tt For\:200\:s\:train\:travel\:for\: constant\:velocity \\ \\ \tt Let\:distance\:travelled\:in\:this\:uniform\:motion\:be\:{s}_{2}\\ \\ \tt {s}_{2} = vt \\ \\ \tt \longmapsto {s}_{2} = 20 × 200 \\ \\ \tt \longmapsto {s}_{2} = 4000 \: m \\ \\ \boxed{\bold{\underline{\red{\tt The\:distance\:travelled\:in\:uniform\:motion\:is\:4000\:m}}}} \\ \\ \tt From\:this\:velocity\:breaks\:are\: applied \\ \\ \tt Now \:Initial\:velocity = 20\:m\:{s}^{-1} \\ \tt Final\:velocity\:(v) = 0 \\ \tt Time\:(t) = 50\:s \\ \\ \tt Using\:the\:same\:formula\\ \\ \tt v = u + at \\ \\ \tt \longmapsto 0 = 20 + a × 50 \\ \\ \tt \longmapsto 50a = -20 \\ \\ \tt \longmapsto a = - \frac{2}{5} \\ \\ \tt \longmapsto a = - 0.4\:m\:{s}^{-2} \\ \\ \boxed{\bold{\underline{\green{\tt The\: retardation\:in\:last\:50\:s\:is\:0.4\:m\:{s}^{-2}}}}} \\ \\ \tt Let\:distance\:travelled\:in\: restarting\:motion\:be\:{s}_{3} \\ \\ \tt \longmapsto {s}_{3} = ut + \frac{1}{2}a{t}^{2} \\ \\ \tt \longmapsto {s}_{3} = 20 × 50 + \frac{1}{2} × 0.4 × 50 × 50 \\ \\ \tt \longmapsto {s}_{3} = 100 + 500 \\ \\ \tt \longmapsto {s}_{3} = 600\:m \\ \\ \boxed{\bold{\underline{\red{\tt Distance\:travelled\:in\:retarding\:motion\:is\:6000\:m }}}}\\ \\ \tt So\:total\:distance\:travelled\:be\:s \\ \\ \tt s = {s}_{1} + {s}_{2} + {s}_{3} \\ \\ \tt \longmapsto s = 20 + 4000 + 600 \\ \\ \tt \longmapsto s = 4620\:m \\ \\ \boxed{\bold{\underline{\green{\tt The\:total\:distance\:travelled\:is\:4620\:m}}}} \\ \\ \tt Total\:distanec\:(s) = 4620 \: m \\ \tt Total\:time = 10 + 200 + 50 = 260\:s \\ \\ \tt Average\:velocity({v}_{arg} = \frac{s}{T} \\ \\ \tt \longmapsto {v}_{arg} = \frac{4620}{260} \\ \\ \tt \longmapsto {v}_{arg} = 17.76\:m\:{s}^{-1} \\ \\ \boxed{\bold{\underline{\green{\tt The\:average\:velocity\:of\:train\:is\:17.76\:m\:{s}^{-1} }}}} \\ \\ \huge{\underline{\underline{\orange{\mathfrak{Niranjan}}}}}$

Answer 2

Explanation:

The plasma membrane, also called the cell membrane, is the membrane found in all cells that separates the interior of the cell from the outside environment. ... The plasma membrane consists of a lipid bilayer that is semipermeable. The plasma membrane regulates the transport of materials entering and exiting the cell.