If an electron is accelerated from rest through a potential difference of 9.9 kV

Question

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? m_e = 9.11 times 10^-31 kg; q_e = 1.6 times 10^-19 C 5.9 times 10^7 m/s 4.9 times 10^7 m/s 3.9 times 10^7 m/s 2.9 times 10^7 m/s 1.9 times 10^7 m/s If each C_1 is 6.9 mu F and each C_2 is 4.6 mu F, calculate the equivalent capacitance of the adjoining circuit. 1.2 mu F 2.3 mu F 3.4 mu F 4.5 mu F 5.6 mu F For the circuit shown in the figure, the current in the 8-Ohm resistor is 0.50 A, and all quantities are accurate to 2 significant figures. What is the current in the 2-Ohm resistor? 2.2 A 0.75 A 4.5 A 9.5 A 6.4 A

Explanation / Answer

Q1.

eV=(1/2)mv^2

v = sqrt( 2*QV/m)

=> v = sqrt((2*1.6*10^-19*9.9*10^3)/(9.1*10^-31))

=> v = 5.9*10^7 m/s

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