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Question-14144

Question Number 14144 by tawa tawa last updated on 28/May/17 Answered by ajfour last updated on 28/May/17 $$\mathrm{69}.\:\:\:\:\:\:\boldsymbol{{I}}_{{b}} =\mathrm{30}{A}\:\:\:\:\:{direction}\:\left(−\boldsymbol{{z}}\right) \\ $$$$\mathrm{70}.\:\:\:\:\boldsymbol{{B}}_{\mathrm{0}} =\frac{\pi}{\mathrm{2}}×\mathrm{10}^{−\mathrm{7}} \:{T}\:. \\ $$…

Question-14079

Question Number 14079 by Tinkutara last updated on 27/May/17 Commented by ajfour last updated on 27/May/17 $${tbreshold}\:{frequency}\:= \\ $$$$\:\:\:\:\:\:\:\left({speed}\:{of}\:{e}.{m}.\:{wave}\right)/\left({threshold}\:{wavelength}\right) \\ $$$$\boldsymbol{\nu}_{\mathrm{0}} =\frac{\boldsymbol{{c}}}{\boldsymbol{\lambda}_{\mathrm{0}} }\:=\frac{\left(\mathrm{3}×\mathrm{10}^{\mathrm{8}} {m}/{s}\right)}{\left(\mathrm{6800}×\mathrm{10}^{−\mathrm{10}} {m}\right)}…

Calculate-the-maximum-number-of-orders-vissible-with-a-diffraction-grating-of-500-lines-per-milimitres-using-light-of-wavelenght-600nm-

Question Number 14025 by tawa tawa last updated on 26/May/17 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{orders}\:\mathrm{vissible}\:\mathrm{with}\:\mathrm{a}\:\mathrm{diffraction}\:\mathrm{grating} \\ $$$$\mathrm{of}\:\mathrm{500}\:\mathrm{lines}\:\mathrm{per}\:\mathrm{milimitres}\:\:\mathrm{using}\:\mathrm{light}\:\mathrm{of}\:\mathrm{wavelenght}\:\mathrm{600nm}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

A-cathode-ray-beam-is-bent-in-a-circle-of-radius-2cm-by-uniform-field-with-B-4-5-10-3-T-What-is-the-speed-of-the-electrons-

Question Number 13965 by tawa tawa last updated on 25/May/17 $$\mathrm{A}\:\mathrm{cathode}\:\mathrm{ray}\:\mathrm{beam}\:\mathrm{is}\:\mathrm{bent}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\:\mathrm{2cm}\:\mathrm{by}\:\mathrm{uniform}\:\mathrm{field}\:\mathrm{with} \\ $$$$\mathrm{B}\:=\:\mathrm{4}.\mathrm{5}\:×\:\mathrm{10}^{−\mathrm{3}} \:\mathrm{T}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{electrons}\:??. \\ $$ Answered by sandy_suhendra last updated on 26/May/17 $$\mathrm{mass}\:\mathrm{of}\:\mathrm{electron}\:=\:\mathrm{m}\:=\:\mathrm{9}.\mathrm{1}×\mathrm{10}^{−\mathrm{31}} \:\mathrm{kg}…