Answer:
a) a = 5.03x10Ā¹Ā³ m/sĀ²
b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]
Explanation: Ā Ā
a) The acceleration of the positron can be found as follows:
[tex] F = q*E [/tex] Ā Ā (1)
Also,
[tex] F = ma [/tex] Ā Ā (2)
By entering equation (1) into (2), we have:
[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]
Where:
F: is the electric force
m: is the particle's mass = 9.1x10ā»Ā³Ā¹ kg
q: is the charge of the positron = 1.6x10ā»Ā¹ā¹ C Ā Ā
E: is the electric field = 286 N/C
[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]
b) The positron's speed can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed =?
[tex]V_{0}[/tex]: is the initial speed =0
t: is the time = 8.70x10ā»ā¹ s
[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]
I hope it helps you!
