A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40
103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.50 cm horizontally.
1 answer:
Answer:
86.5 ns
Step-by-step explanation:
The speed in the original direction (horizontally) is unchanged by the vertical force the field exerts. The travel time is ...
time = distance/speed = (4.5×10^-2 m)/(5.20×10^5 m/s) = 8.65×10^-8 s
_____
An engineer would express this time using the SI prefix nano- for 10^-9. The time is 86.5 ns.
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