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CaA Particle carrying charge of 2e falls through a Potential difference of 3.ovCalculate the energy acquired by it? |
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Answer» ANSWER: Charge on particle = 2e Potential difference = 3.0 V Energy? By applying FORMULA 1/2 mv²= QV K. E = 2e×3 = 2×1.6×10^-19×3 Energy = 9.6×10^-19J Hope you get it CLEARLY, Thanks Explanation: |
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