Question Number 183010 by sciencestudent last updated on 18/Dec/22 $${what}\:{is}\:{the}\:{formula}\:{between}\:{actual}\:{depth} \\ $$$${apparent}\:{depth}\:{and}\:{refraction}\:{index} \\ $$$${that}\:{an}\:{object}\:{is}\:{in}\:{a}\:{rarer}\:{medium}? \\ $$ Commented by sciencestudent last updated on 18/Dec/22 $$? \\…
Question Number 51938 by hassentimol last updated on 01/Jan/19 $$\mathrm{To}\:\mathrm{which}\:\mathrm{interaction}\:\mathrm{among}\:\mathrm{the}\:\mathrm{four} \\ $$$$\mathrm{fundamental}\:\mathrm{interactions}\:\mathrm{is}\:\mathrm{linked} \\ $$$$\mathrm{the}\:\mathrm{combustion}\:?\:\mathrm{the}\:\mathrm{fission}\:? \\ $$$$ \\ $$$$\mathrm{Thanks}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated…
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Question Number 182954 by Gamil last updated on 17/Dec/22 $$\:\boldsymbol{{I}}=\int\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}\:\sqrt{\boldsymbol{{x}}}\:−\boldsymbol{{x}}^{\mathrm{2}} }}\:\boldsymbol{{dx}}\: \\ $$$$\:\:\boldsymbol{{I}}=\int\:\frac{\mathrm{1}}{\:\:\centerdot\:\:\sqrt{\boldsymbol{{x}}\left(\sqrt{\boldsymbol{{x}}}\:−\boldsymbol{{x}}\right)}}\:\boldsymbol{{dx}} \\ $$$$\:\:\boldsymbol{{I}}=\int\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}}\:\centerdot\:\sqrt{\sqrt{\boldsymbol{{x}}}\:−\boldsymbol{{x}}}\:}\:×\frac{\mathrm{2}}{\mathrm{2}}\:\boldsymbol{{dx}} \\ $$$$\:\boldsymbol{{I}}=\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\sqrt{\boldsymbol{{x}}}\:−\mathrm{4}\boldsymbol{{x}}}}\:\centerdot\:\frac{\mathrm{2}}{\:\sqrt{\boldsymbol{{x}}}}\:\boldsymbol{{dx}} \\ $$$$\:\:\boldsymbol{{I}}=\mathrm{2}\int\:\frac{\:}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\:\sqrt{\boldsymbol{{x}}}\right)^{\mathrm{2}} }}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}}}\:\boldsymbol{{dx}}\: \\ $$$$\:\:\boldsymbol{{I}}=\:−\mathrm{2}\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\sqrt{\boldsymbol{{x}}}\right)^{\mathrm{2}} }}\:\:\centerdot\:\boldsymbol{{d}}\left(\mathrm{1}−\mathrm{2}\sqrt{\boldsymbol{{x}}}\right) \\ $$$$\:\:\boldsymbol{{I}}=\:−\mathrm{2}\boldsymbol{{sin}}^{−\mathrm{1}}…