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In a high-energy physics experiment, a subnuclear particle moves in a circular arc of 4.35

In a high-energy physics experiment, a subnuclear particle moves in a circular arc of 4.35

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Answer #1

KE = 0.5*m*v^2

v = sqrt(2K/m)


Fb = Fc


B*q*v = m*v^2/r

B*q = m*v/r

m * v = B*q*r

m*sqrt(2k/m) = B*q*r


squaring on both sides


m^2*2*k/m = B^2*q^2*r^2

m*2*k = (B*q*r)^2

m = (B*q*r)^2/2k


m = (2.7e-2*1.6e-19*4.35e-2)^2/(2*9.4e-17)

m = 1.88*10^-28 kg


the particle muon

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Answer #2

Velocity of particle entering field: KE-0.5 mm2 v root of [ (2 x KE)/m].... 1) Remember Fine Bream Tubel putting in (1) into

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