A charged particle moves in a uniform electric field and travels a distance of 1.0 cm along the field. If its charge-to-mass ratio is such that acceleration is a = qE/m = 9.0×10^15 m/s^2, what is its final speed after traveling 1.0 cm (assume it starts from rest)?

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Study for the UCF PHY2054 General Physics Exam. Use flashcards and multiple-choice questions complete with hints and explanations. Boost your understanding and get exam-ready!

Multiple Choice

A charged particle moves in a uniform electric field and travels a distance of 1.0 cm along the field. If its charge-to-mass ratio is such that acceleration is a = qE/m = 9.0×10^15 m/s^2, what is its final speed after traveling 1.0 cm (assume it starts from rest)?

Explanation:
Constant acceleration with starting from rest links final speed to distance via v^2 = v0^2 + 2 a s. Since v0 = 0, the final speed is v = sqrt(2 a s). Plugging in s = 0.01 m and a = 9.0×10^15 m/s^2 gives 2 a s = 2 × 9.0×10^15 × 0.01 = 1.8×10^14. Taking the square root, v ≈ sqrt(1.8) × 10^7 ≈ 1.34×10^7 m/s. With the given sig figs, this is about 1.33×10^7 m/s. So the final speed is roughly 1.33×10^7 m/s.

Constant acceleration with starting from rest links final speed to distance via v^2 = v0^2 + 2 a s. Since v0 = 0, the final speed is v = sqrt(2 a s).

Plugging in s = 0.01 m and a = 9.0×10^15 m/s^2 gives 2 a s = 2 × 9.0×10^15 × 0.01 = 1.8×10^14. Taking the square root, v ≈ sqrt(1.8) × 10^7 ≈ 1.34×10^7 m/s. With the given sig figs, this is about 1.33×10^7 m/s.

So the final speed is roughly 1.33×10^7 m/s.

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