JAMB Physics 2014 Objective — Question 26
Question 26 of 50 from the Joint Admissions and Matriculation Board (JAMB) Physics 2014 Objective paper, with the correct answer and a full explanation.
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The distance between two successive crests of a wave is 15 cm and the velocity is 300 ms⁻¹. Calculate the frequency.
- A. 4.5 ×10³Hz
- B. 2.0 ×10³HzCorrect
- C. 4.5 ×10²Hz
- D. 2.0 ×10²Hz
Explanation
F=V/λ = 300/0.15 = 2000Hz = 2.0×10³Hz.
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