An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
To find how many electrons flow through an electrical device delivering 15.0 A for 30 seconds, we first calculate the total charge using the formula Q = I × t , resulting in 450 C. Next, we divide this charge by the charge of an electron (about 1.6 × 1 0 − 19 C ), which gives us approximately 2.81 × 1 0 21 electrons. Thus, about 2.81 sextillion electrons flow through the device.
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Multiply the coefficients outside the radicals: − 2 × 5 = − 10 .
Multiply the expressions inside the radicals: 20 k × 8 k 3 = 160 k 4 .
Simplify the radical: 160 k 4 = 4 k 2 10 .
Multiply by the coefficient: − 10 × 4 k 2 10 = − 40 k 2 10 . The final answer is − 40 k 2 10 .
Explanation
Initial Analysis We are given the expression ( − 2 20 k ) ( 5 8 k 3 ) and asked to simplify it. First, we multiply the coefficients outside the radicals and the expressions inside the radicals separately.
Multiplying Coefficients and Radicals Multiplying the coefficients, we have − 2 × 5 = − 10 . Multiplying the expressions inside the radicals, we get 20 k × 8 k 3 = 20 k × 8 k 3 = 160 k 4 .
Simplifying the Radical Combining these results, we have − 10 160 k 4 . Now, we simplify the radical. We can factor 160 as 160 = 16 × 10 . Also, k 4 is a perfect square since k 4 = ( k 2 ) 2 . Thus, we can rewrite the radical as 160 k 4 = 16 × 10 × k 4 = 16 × 10 × k 4 = 4 k 2 10 .
Final Simplification Finally, we multiply this simplified radical by the coefficient we found earlier: − 10 × 4 k 2 10 = − 40 k 2 10 . Therefore, the simplified expression is − 40 k 2 10 .
Examples
Radicals are often used in engineering to calculate things like the natural frequencies of systems, which can help engineers design structures that don't resonate dangerously. Simplifying radical expressions makes these calculations easier to handle. For example, when analyzing the vibration of a bridge, engineers might encounter expressions involving square roots of lengths and masses. Simplifying these expressions helps them determine the bridge's natural frequencies and ensure its stability under various loads.
