Asked by koraile
                An electron with a speed of 1.3 × 107 m/s moves horizontally into a region where a constant vertical force of 4.4 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 29 mm horizontally.
            
            
        Answers
                    Answered by
            drwls
            
    There is vertical acceleration 
ay = F/m = 4.83*10^14 m/s^2
The time it takes to move 29 mm = 0.029 m horizontally is
T = 0.029 m/1.3*10^7s = 2.23*10^-9 s
The vertical distance moved is
Y = (1/2)*ay*T^2 = 2.4*10-3 m = 2.4 mm
    
ay = F/m = 4.83*10^14 m/s^2
The time it takes to move 29 mm = 0.029 m horizontally is
T = 0.029 m/1.3*10^7s = 2.23*10^-9 s
The vertical distance moved is
Y = (1/2)*ay*T^2 = 2.4*10-3 m = 2.4 mm
                    Answered by
            koraile
            
    its giving me a wrong answear can u check again plz
    
                    Answered by
            koraile
            
    it is giving me 0.0024m as a wrong answer
    
                    Answered by
            joe
            
    THANKS!!!
    
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