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Question Number 27128 by NECx last updated on 02/Jan/18 $${A}\:{body}\:{resting}\:{on}\:{a}\:{rough} \\ $$$${horizontal}\:{plane}\:{require}\:{a}\:{pull}\:{of} \\ $$$$\mathrm{18}{N}\:{inclined}\:{at}\:\mathrm{30}°\:{to}\:{the}\:{plane} \\ $$$${first}\:{to}\:{move}\:{it}.{It}\:{was}\:{found} \\ $$$${that}\:{a}\:{push}\:{of}\:\mathrm{22}{N}\:{inclined}\:{at}\:\mathrm{30}° \\ $$$${to}\:{the}\:{plane}\:{just}\:{moved}\:{the}\:{body}. \\ $$$${Determine}\:{the}\:{weight}\:{and}\: \\ $$$${coefficient}\:{of}\:{friction}. \\…
Question Number 27117 by Tinkutara last updated on 02/Jan/18 Answered by ajfour last updated on 02/Jan/18 $$\frac{{dv}}{{dt}}=−\left(\frac{{v}^{\mathrm{2}} }{{r}}\right)\:,\:\:\left({becoz}\:\:{a}_{{t}} =−{a}_{{r}} =−\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$$\int_{\mathrm{10}} ^{\:\:\mathrm{5}} \frac{{dv}}{{v}^{\mathrm{2}}…
Question Number 27111 by Tinkutara last updated on 02/Jan/18 Answered by mrW1 last updated on 02/Jan/18 $${T}={tension}\:{in}\:{string} \\ $$$${a}_{{A}} =\mathrm{2}{a}_{{B}} \\ $$$${m}_{{B}} {g}−\mathrm{2}{T}={m}_{{B}} {a}_{{B}} \\…
Question Number 27102 by Tinkutara last updated on 02/Jan/18 Answered by ajfour last updated on 02/Jan/18 $${let}\:{reaction}\:{force}\:{be}\:{N}. \\ $$$${N}\mathrm{sin}\:\theta={mA}\:\:\:\left({A}:\:{acc}.{of}\:{shell}\right) \\ $$$${pseudo}\:{force}\:{on}\:{particle}\:{is}\:{mA}. \\ $$$${so}\:\:\frac{{N}}{{F}_{{pseudo}} }\:=\:\frac{{mA}\mathrm{cosec}\:\theta}{{mA}}\:=\mathrm{cosec}\:\theta\:. \\…
Question Number 27101 by Tinkutara last updated on 02/Jan/18 Answered by mrW1 last updated on 02/Jan/18 $${y}={ut}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{u}}{{g}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} =\frac{\mathrm{2}{au}^{\mathrm{2}} }{{g}^{\mathrm{2}} }…
Question Number 27075 by Tinkutara last updated on 01/Jan/18 $$\mathrm{Laws}\:\mathrm{of}\:\mathrm{Motion}\:\mathrm{question}\:\mathrm{at} \\ $$$$\mathrm{ibb}.\mathrm{co}/\mathrm{cqq1NG} \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{uploading}\:\mathrm{here}\:\mathrm{but}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{get}\:\mathrm{uploaded}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27044 by Tinkutara last updated on 01/Jan/18 Answered by mrW1 last updated on 02/Jan/18 $$\alpha=\mathrm{30}° \\ $$$${x}={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$${y}={u}\:\mathrm{sin}\:\theta\:{t}\:−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${by}\:{landing}: \\…
Question Number 27033 by Tinkutara last updated on 01/Jan/18 Answered by mrW1 last updated on 01/Jan/18 $$\left(\frac{{L}}{\mathrm{2}{h}}\right)^{\mathrm{2}} =\frac{{H}}{{H}−{h}}=\frac{{H}}{{H}−\frac{\mathrm{3}{H}}{\mathrm{4}}}=\mathrm{4} \\ $$$$\Rightarrow{L}=\mathrm{4}{h} \\ $$ Commented by mrW1…
Question Number 26982 by Tinkutara last updated on 31/Dec/17 Answered by mrW1 last updated on 31/Dec/17 $${Q}\mathrm{1} \\ $$$${m}_{{B}} {a}=\mu_{{S}} {m}_{{B}} {g} \\ $$$$\Rightarrow{a}=\mu_{{S}} {g}=\mathrm{0}.\mathrm{6}×\mathrm{10}=\mathrm{6}\:{m}/{s}^{\mathrm{2}}…