Answered: A proton (q = 1.60 x 10-1⁹ C, m = 1.67… | bartleby

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**Problem Statement:**

A proton (\(q = 1.60 \times 10^{-19}\) C, \(m = 1.67 \times 10^{-27}\) kg), is released from rest in a 6 N/C uniform electric field. How long (in \(\mu\)s) does it take to move 7 m from its starting point?

**Solution Space:**

- Answer: \(\_\_\_\_\_\) \(\mu\)s

- Correct Answer: 156.0582

- Margin of error: +/- 1%

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$P r o t o n space c h a r g e space q equals 1.6 cross times 10 to the power of negative 19 end exponent space C m a s s space m equals 1.67 cross times 10 to the power of negative 27 end exponent space k g E l e c t r i c f i e l d space E equals 6 space N divided by C D i s tan c e space d equals 7 space m$

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