A 1.50 V battery supplies 0.443 W of power to a small flashlight for 16.2 min. (a) How much charge does it move? C (b) How many electrons must move to carry this charge? electrons
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Given: Potential (V) = 1.50 V Power (P) = 0.443 W Time (t) = 16.2 min = 16.2 * 60 s = 972 s We know that power (P) = potential (V) * current (I) Therefore, current (I) = P / V Substitute the values: I = 0.443 W / 1.50 V I = 0.295 A Now, we know that charge (Q) Show more…
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