Question Number 27564 by Tinkutara last updated on 09/Jan/18 Commented by Tinkutara last updated on 09/Jan/18 Commented by mrW2 last updated on 12/Jan/18 Commented by…
Question Number 27435 by NECx last updated on 06/Jan/18 $${An}\:{object}\:{has}\:{a}\:{constant} \\ $$$${acceleration}\:{a}=\mathrm{4}{ms}^{−\mathrm{2}} .{Its}\:{velocity} \\ $$$${is}\:\mathrm{1}{m}/{s}\:{at}\:{t}=\mathrm{0},{when}\:{it}\:{is}\:{at}\:{x}=\mathrm{7}{m}. \\ $$$${How}\:{fast}\:{is}\:{it}\:{at}\:{x}=\mathrm{8}{m}?{At}\:{what} \\ $$$${time}\:{is}\:{this}? \\ $$ Answered by mrW1 last…
Question Number 27436 by NECx last updated on 06/Jan/18 $${Car}\:{A}\:{is}\:{moving}\:{at}\:{a}\:{speed}\:{of}\: \\ $$$$\mathrm{45}{km}/{h}\:{towards}\:{car}\:{B}\:{which}\:{is} \\ $$$${moving}\:{towards}\:{car}\:{A}\:{at}\:{a}\:{speed}\: \\ $$$${of}\:\mathrm{55}{km}/{h}.{If}\:{the}\:{two}\:{cars}\:{where} \\ $$$${initially}\:{seperated}\:{at}\:{a}\:{distance} \\ $$$${of}\:\mathrm{150}{m}.{Determine}\:{how}\:{long}\:{it}\: \\ $$$${will}\:{take}\:{the}\:{two}\:{cars}\:{to}\:{meet}. \\ $$ Commented…
Question Number 27429 by Tinkutara last updated on 06/Jan/18 Answered by mrW1 last updated on 07/Jan/18 $${let}'{s}\:{see}\:{point}\:\mathrm{2}\:{as}\:{example}, \\ $$$$\frac{{v}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} }{{v}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} }={e} \\ $$$$\Rightarrow\:{v}_{\mathrm{2}}…
Question Number 27326 by NECx last updated on 05/Jan/18 $${What}\:{is}\:{the}\:{relationship}\:{between} \\ $$$${the}\:{centre}\:{of}\:{gravity}\:{and}\:{the}\:{centre}\:{of} \\ $$$${mass}? \\ $$ Answered by prakash jain last updated on 05/Jan/18 $$\mathrm{Center}\:\mathrm{of}\:\mathrm{gravity}\:\mathrm{and}\:\mathrm{center}\:\mathrm{of}\:\mathrm{mass}…
Question Number 27303 by Tinkutara last updated on 04/Jan/18 Answered by mrW1 last updated on 05/Jan/18 $${a}={a}_{//} =\frac{{g}}{\mathrm{2}}\:\left({given}\right) \\ $$$${a}_{\bot} ={a}\:\mathrm{sin}\:\theta=\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${mg}\:\mathrm{cos}\:\theta−{N}={ma}_{\bot} =\frac{{mg}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\…
Question Number 27202 by Tinkutara last updated on 03/Jan/18 Answered by mrW1 last updated on 03/Jan/18 $${N}={normal}\:{force}\:{between}\:{m}\:{and}\:{M} \\ $$$${T}={tension}\:{in}\:{thread} \\ $$$${N}={mg}\:\mathrm{cos}\:\theta \\ $$$${T}={N}\:\mathrm{sin}\:\theta={mg}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta=\mathrm{3}×\mathrm{10}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{15}\:{N} \\ $$…
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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}}…