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an-object-of-12cm-long-is-placed-at-the-centre-of-curvature-of-concave-mirror-whose-focal-length-is-40cm-the-location-of-the-image-formed-is-

Question Number 132899 by aurpeyz last updated on 17/Feb/21 $${an}\:{object}\:{of}\:\mathrm{12}{cm}\:{long}\:{is}\:{placed}\:{at}\:{the}\: \\ $$$${centre}\:{of}\:{curvature}\:{of}\:{concave}\:{mirror} \\ $$$${whose}\:{focal}\:{length}\:{is}\:\mathrm{40}{cm}.\:{the}\:{location} \\ $$$${of}\:{the}\:{image}\:{formed}\:{is}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

cot-142-5-

Question Number 132880 by faysal last updated on 17/Feb/21 $${cot}\left(\mathrm{142}.\mathrm{5}°\right)=? \\ $$ Commented by bramlexs22 last updated on 17/Feb/21 $$\mathrm{285}°=\:\mathrm{2}×\mathrm{142}.\mathrm{5} \\ $$$$\mathrm{285}°=\mathrm{270}°+\mathrm{15}° \\ $$$$\mathrm{cot}\:\mathrm{285}°\:=\:\mathrm{cot}\:\mathrm{285}° \\…

linear-magnification-for-convex-mirror-image-height-object-height-image-distance-object-distance-is-this-formula-correct-what-is-the-meaning-of-the-negative-sign-

Question Number 132872 by aurpeyz last updated on 17/Feb/21 $${linear}\:{magnification}\:{for}\:{convex} \\ $$$${mirror} \\ $$$$=\frac{{image}\:{height}}{{object}\:{height}}=\frac{−{image}\:{distance}}{{object}\:{distance}} \\ $$$$ \\ $$$${is}\:{this}\:{formula}\:{correct}?\:{what}\:{is}\:{the} \\ $$$${meaning}\:{of}\:{the}\:{negative}\:{sign}? \\ $$ Terms of Service…