Question Number 33272 by NECx last updated on 14/Apr/18 $${The}\:\boldsymbol{{resistance}}\:\boldsymbol{{R}}\:\boldsymbol{{of}}\:\boldsymbol{{an}}\: \\ $$$$\boldsymbol{{unknown}}\:\boldsymbol{{resistor}}\:\boldsymbol{{is}}\:\boldsymbol{{found}}\:\boldsymbol{{by}} \\ $$$$\boldsymbol{{measuring}}\:\boldsymbol{{the}}\:\boldsymbol{{potential}} \\ $$$$\boldsymbol{{difference}}\:\boldsymbol{{V}}\:\boldsymbol{{across}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{resistor}}\:\boldsymbol{{and}}\:\boldsymbol{{the}}\:\boldsymbol{{current}}\:\boldsymbol{{I}}\:\boldsymbol{{through}} \\ $$$$\boldsymbol{{it}}\:\boldsymbol{{and}}\:\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}}\:\boldsymbol{{R}}=\frac{{V}}{{I}}. \\ $$$$\boldsymbol{{The}}\:\boldsymbol{{voltmeter}}\:\boldsymbol{{reading}}\:\boldsymbol{{has}}\:\boldsymbol{{a}}\:\mathrm{3\%} \\ $$$$\boldsymbol{{uncertainty}}\:\boldsymbol{{and}}\:\boldsymbol{{the}}\:\boldsymbol{{ammeter}} \\…
Question Number 33239 by Tinkutara last updated on 13/Apr/18 Answered by ajfour last updated on 14/Apr/18 $$\mathrm{2}\left[{mu}\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}\right)\right]=\left(\mathrm{2}{m}\right)\left(\frac{{u}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\Rightarrow\:\:\alpha\:=\:\mathrm{120}°\:. \\ $$ Commented by Tinkutara last…
Question Number 33214 by Tinkutara last updated on 13/Apr/18 Commented by Tinkutara last updated on 13/Apr/18 Commented by Tinkutara last updated on 15/Apr/18 But if we consider the torque of mg instead of ma it still gives approximately the same answer. Which one is conceptually correct? Commented…
Question Number 164256 by ajfour last updated on 15/Jan/22 Answered by ajfour last updated on 16/Jan/22 $${A}\:\&\:{B}\:{represent}\:{points}\:{and}\:{also} \\ $$$${forces}\:{at}\:{those}\:{points}\:{respectively}. \\ $$$${q}=−\left(\mathrm{2}{a}\mathrm{cos}\:\theta−{p}\right) \\ $$$$\mathrm{2}{p}=\mathrm{tan}\:\alpha \\ $$$$\mathrm{2}{q}=−\mathrm{tan}\:\beta…
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Question Number 98575 by aurpeyz last updated on 14/Jun/20 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{force}\:\mathrm{F}\:\mathrm{needed}\:\mathrm{to}\:\mathrm{punch}\:\mathrm{at}\:\mathrm{1}.\mathrm{46cm}\:\mathrm{diametre}\:\mathrm{hole}\:\mathrm{in}\:\mathrm{a}\:\mathrm{steel}\:\mathrm{plate}\:\mathrm{1}.\mathrm{27cm}\:\mathrm{tjick}.\:\mathrm{tbe}\:\mathrm{ultimate}\:\mathrm{shear}\:\mathrm{stress}\:\mathrm{of}\:\mathrm{the}\:\mathrm{steel}\:\mathrm{is}\:\mathrm{3}.\mathrm{45}×\mathrm{10}^{\mathrm{8N}/\mathrm{m}^{\mathrm{2}} } \\ $$ Commented by aurpeyz last updated on 14/Jun/20 $$\mathrm{pls}\:\mathrm{he}\:\mathrm{me}\:\mathrm{with}\:\mathrm{it}? \\ $$ Commented by…
Question Number 98568 by aurpeyz last updated on 14/Jun/20 Commented by aurpeyz last updated on 14/Jun/20 $$\mathrm{pls}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{question}\:\mathrm{13}.\mathrm{9} \\ $$ Answered by mr W last updated…
Question Number 164073 by mr W last updated on 13/Jan/22 Commented by mr W last updated on 13/Jan/22 $${are}\:{there}\:{other}\:{equilibrium}\:{positions} \\ $$$${for}\:{the}\:{rod}? \\ $$ Answered by…
Question Number 163993 by mathls last updated on 12/Jan/22 Commented by mathls last updated on 12/Jan/22 $${please}\:{help} \\ $$ Commented by mr W last updated…
Question Number 163912 by ajfour last updated on 11/Jan/22 Commented by mr W last updated on 11/Jan/22 $${nice}\:{question},\:{but}\:{quite}\:{hard} \\ $$ Terms of Service Privacy Policy…