Respuesta :
Answer:
a) 7.35 x 10¹³ m/s²
b) 5.03 x 10⁻⁸ sec
c) 9.3 cm
d) 6.23 x 10⁻¹⁸ J
Explanation:
E = magnitude of electric field = 418 N/C
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of the electron = 9.1 x 10⁻³¹ kg
a)
acceleration of the electron is given as
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}[/tex]
a = 7.35 x 10¹³ m/s²
b)
v = final velocity of the electron = 3.70 x 10⁶ m/s
v₀ = initial velocity of the electron = 0 m/s
t = time taken
Using the equation
v = v₀ + at
3.70 x 10⁶ = 0 + (7.35 x 10¹³) t
t = 5.03 x 10⁻⁸ sec
c)
d = distance traveled by the electron
using the equation
d = v₀ t + (0.5) at²
d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²
d = 0.093 m
d = 9.3 cm
d)
Kinetic energy of the electron is given as
KE = (0.5) m v²
KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²
KE = 6.23 x 10⁻¹⁸ J
