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Question-55368




Question Number 55368 by Tawa1 last updated on 22/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
(800−160)^2 =(800)^2 −2as_1   2as_1 =(800)^2 −(640)^2   2as_1 =1440×160...(1)  (640−160)^2 =(640)^2 −2(a+((a×50)/(100)))s_2   2×((3a)/2)×s_2 =(640)^2 −(480)^2 =(1120×160  ((2as_1 )/(2×((3a)/2)×s_2 ))=((1440×160)/(1120×160))  ((2s_1 )/(3s_2 ))=((144)/(112))  (s_1 /s_2 )=((36)/(28))×(3/2)  (s_1 /s_2 )=((9×3)/(7×2))=((27)/(14))
$$\left(\mathrm{800}−\mathrm{160}\right)^{\mathrm{2}} =\left(\mathrm{800}\right)^{\mathrm{2}} −\mathrm{2}{as}_{\mathrm{1}} \\ $$$$\mathrm{2}{as}_{\mathrm{1}} =\left(\mathrm{800}\right)^{\mathrm{2}} −\left(\mathrm{640}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{as}_{\mathrm{1}} =\mathrm{1440}×\mathrm{160}…\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{640}−\mathrm{160}\right)^{\mathrm{2}} =\left(\mathrm{640}\right)^{\mathrm{2}} −\mathrm{2}\left({a}+\frac{{a}×\mathrm{50}}{\mathrm{100}}\right){s}_{\mathrm{2}} \\ $$$$\mathrm{2}×\frac{\mathrm{3}{a}}{\mathrm{2}}×{s}_{\mathrm{2}} =\left(\mathrm{640}\right)^{\mathrm{2}} −\left(\mathrm{480}\right)^{\mathrm{2}} =\left(\mathrm{1120}×\mathrm{160}\right. \\ $$$$\frac{\mathrm{2}{as}_{\mathrm{1}} }{\mathrm{2}×\frac{\mathrm{3}{a}}{\mathrm{2}}×{s}_{\mathrm{2}} }=\frac{\mathrm{1440}×\mathrm{160}}{\mathrm{1120}×\mathrm{160}} \\ $$$$\frac{\mathrm{2}{s}_{\mathrm{1}} }{\mathrm{3}{s}_{\mathrm{2}} }=\frac{\mathrm{144}}{\mathrm{112}} \\ $$$$\frac{{s}_{\mathrm{1}} }{{s}_{\mathrm{2}} }=\frac{\mathrm{36}}{\mathrm{28}}×\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{s}_{\mathrm{1}} }{{s}_{\mathrm{2}} }=\frac{\mathrm{9}×\mathrm{3}}{\mathrm{7}×\mathrm{2}}=\frac{\mathrm{27}}{\mathrm{14}} \\ $$
Commented by Tawa1 last updated on 23/Feb/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19
pls recheck question 55375...to find point if application  of tangential force...to me it is not clear...
$${pls}\:{recheck}\:{question}\:\mathrm{55375}…{to}\:{find}\:{point}\:{if}\:{application} \\ $$$${of}\:{tangential}\:{force}…{to}\:{me}\:{it}\:{is}\:{not}\:{clear}… \\ $$

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