%3Cp%3ECathode%20Ray%26nbsp%3BTube%26nbsp%3B%26amp%3B%26nbsp%3BMinimum%26nbsp%3B
ID: 2270659 • Letter: #
Question
%3Cp%3ECathode%20Ray%26nbsp%3BTube%26nbsp%3B%26amp%3B%26nbsp%3BMinimum%26nbsp%3BSpeed%26nbsp%3Bof%26nbsp%3BElectrons%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3E%3Cimg%26nbsp%3Bclass%3D%22decoded%22%26nbsp%3Balt%3D%22http%3A%2F%2Fgyazo.com%2Fd497a2e9bb0e94dc595224c7b2087936.png%22%26nbsp%3Bsrc%3D%22http%3A%2F%2Fgyazo.com%2Fd497a2e9bb0e94dc595224c7b2087936.png%22%3E%3C%2Fp%3E%3Cp%3EThanks%26nbsp%3Bto%26nbsp%3Bwhoever%26nbsp%3Bcan%26nbsp%3Bprovide%26nbsp%3Ba%26nbsp%3Bgood%26nbsp%3Banswer%26nbsp%3Bthat%26nbsp%3Bis%26nbsp%3Bnot%26nbsp%3Bonly%26nbsp%3Bcorrect%26nbsp%3Bbut%26nbsp%3Bwill%26nbsp%3Bhelp%26nbsp%3Bme%26nbsp%3Blearn!%3Cbr%3E%3C%2Fp%3EExplanation / Answer
F = Eq= ma
so
a = Eq/m
now
time of travel through distance = L/v
and change in y in this time
= s= 0.5(Eq/m)*(L/V)^2
v= L*sqrt(0.5(Eq/m)/s)
here s =d/2 = 0.001 m
so
v = 0.03*sqrt(0.5*(1.7*10^4*1.6*10^-19/(9.1*10^-31*0.001)
= 3.667 *10^7 m/s
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