Menu Close

Category: Others

Question-198869

Question Number 198869 by Tawa11 last updated on 25/Oct/23 Commented by mr W last updated on 25/Oct/23 $$\mathrm{26}.\mathrm{8}\:\mathrm{cos}\:\mathrm{10}°\neq\mathrm{28}.\mathrm{2}\:\mathrm{cos}\:\mathrm{20}° \\ $$$$\Rightarrow{something}\:{in}\:{question}\:{is}\:{wrong}! \\ $$ Commented by Tawa11…

Question-198860

Question Number 198860 by cherokeesay last updated on 25/Oct/23 Answered by mr W last updated on 25/Oct/23 $$\left.{D}\right)\:{is}\:{correct}. \\ $$$${the}\:{force}\:{moved}\:{the}\:{block}\:{to}\:{a}\:{height} \\ $$$${a}\:\mathrm{sin}\:\theta,\:{the}\:{work}\:{done}\:{is}\:{wa}\:\mathrm{sin}\:\theta. \\ $$$${the}\:{force}\:{pulled}\:{the}\:{spring}\:{to}\:{a}\:{length} \\…

3x-2-1-x-5x-3x-2-x-1-119-18-

Question Number 198566 by tri26112004 last updated on 22/Oct/23 $$\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{{x}}+\frac{\mathrm{5}{x}}{\mathrm{3}{x}^{\mathrm{2}} −{x}−\mathrm{1}}=\frac{\mathrm{119}}{\mathrm{18}} \\ $$ Answered by Rasheed.Sindhi last updated on 22/Oct/23 $$\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{{x}}+\frac{\mathrm{5}{x}}{\mathrm{3}{x}^{\mathrm{2}} −{x}−\mathrm{1}}=\frac{\mathrm{119}}{\mathrm{18}} \\…

The-furier-series-approximation-to-the-forcing-function-is-given-by-f-t-5-1-4-pi-sin120pit-1-sin360pit-2-sin600pit-3-The-transfer-function-for-this-problem-T-s-X

Question Number 198465 by BHOOPENDRA last updated on 20/Oct/23 $${The}\:{furier}\:{series}\:{approximation}\:{to}\: \\ $$$${the}\:{forcing}\:{function}\:{is}\:{given}\:{by}\: \\ $$$${f}\left({t}\right)=\mathrm{5}\left[\mathrm{1}+\frac{\mathrm{4}}{\pi}\left(\frac{}{}\frac{{sin}\mathrm{120}\pi{t}}{\mathrm{1}}+\frac{{sin}\mathrm{360}\pi{t}}{\mathrm{2}}+\frac{{sin}\mathrm{600}\pi{t}}{\mathrm{3}}\right.\right. \\ $$$$\left.\:\left.\:\:\:\:\:+………\right)\right] \\ $$$${The}\:{transfer}\:{function}\:{for}\:{this} \\ $$$${problem}\:\:{T}\left({s}\right)=\frac{{X}\left({s}\right)}{{f}\left({s}\right)}=\frac{\mathrm{1}}{{ms}^{\mathrm{2}} +{cs}+{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{0}.\mathrm{001}{s}+\mathrm{1}} \\ $$$$\mathrm{1}.\:{plot}\:{the}\:{amplitude}\:{spectrum}\:…

Question-198309

Question Number 198309 by BHOOPENDRA last updated on 17/Oct/23 Commented by BHOOPENDRA last updated on 17/Oct/23 $${Sketch}\:{the}\:{partial}\:{auxiliary}\:{view}\:{for}\: \\ $$$${the}\:{auxiliary}\:{surface}\:{object}\:,{given}\: \\ $$$${the}\:{front}\:{and}\:{right}\:{view}. \\ $$ Commented by…

Question-198186

Question Number 198186 by Tawa11 last updated on 13/Oct/23 Answered by mr W last updated on 13/Oct/23 $${y}\left({t}\right)=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi{t}}{{T}}\right) \\ $$$${y}\left(\mathrm{0}.\mathrm{4}\right)=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi×\mathrm{0}.\mathrm{4}}{\mathrm{1}.\mathrm{0}}\right)=\mathrm{0}.\mathrm{09}\:{cm} \\ $$ Commented by Tawa11…