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3 years ago

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An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1

microsecond is 1x10 65). (Express your answers in vector form.) Part 1 (a) What is the average velocity of the electron? Vavg = < > m/s Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER Part 2 (b) If the electron continues to travel at this average velocity, where will it be in another 9 us? 7 = < > m

1 answer:

hram777 [196]3 years ago

5 0

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity = $\frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}$

Average velocity = $< 0, 316000, 146000> m/s$

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d = < 0, 316000, 146000> x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

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