An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Apply the Law of Cosines: a 2 = b 2 + c 2 − 2 b c cos ( A ) .
Substitute the given values: a 2 = 3 2 + 3 2 − 2 ( 3 ) ( 3 ) cos ( 4 π ) .
Simplify using cos ( 4 π ) = 2 2 : a 2 = 18 − 18 ( 2 2 ) = 18 − 9 2 .
Factor the expression: a 2 = 9 ( 2 − 2 ) = 3 2 ( 2 − 2 ) .
3 2 ( 2 − 2 )
Explanation
Problem Analysis We are given an isosceles triangle A BC with A = 4 π and b = c = 3 . We need to find the value of a 2 , where a is the length of the side opposite to angle A .
Applying the Law of Cosines We can use the Law of Cosines to relate the sides and angles of the triangle. The Law of Cosines states: a 2 = b 2 + c 2 − 2 b c cos ( A ) In our case, A = 4 π and b = c = 3 .
Substitution Substitute the given values into the Law of Cosines: a 2 = 3 2 + 3 2 − 2 ( 3 ) ( 3 ) cos ( 4 π ) a 2 = 9 + 9 − 18 cos ( 4 π )
Simplifying the expression We know that cos ( 4 π ) = 2 2 . Substitute this value into the equation: a 2 = 18 − 18 ( 2 2 ) a 2 = 18 − 9 2
Factoring Factor out 9 from the expression: a 2 = 9 ( 2 − 2 ) a 2 = 3 2 ( 2 − 2 )
Final Answer Therefore, the length of a 2 is 3 2 ( 2 − 2 ) .
Examples
The Law of Cosines is a fundamental concept in trigonometry that relates the lengths of the sides of a triangle to the cosine of one of its angles. It's particularly useful in fields like surveying and navigation. For example, surveyors use the Law of Cosines to calculate distances and angles in irregular land plots, where direct measurement is difficult or impossible. Similarly, navigators use it to determine the position and course of a ship or aircraft, especially when GPS data is unavailable or unreliable. Imagine you're designing a triangular garden bed where two sides are each 3 meters long, and the angle between them is 45 degrees. Using the Law of Cosines, you can calculate the length of the third side to determine how much edging you'll need, ensuring a perfectly shaped garden.
The total charge delivered by a device with a current of 15.0 A for 30 seconds is 450 C. This corresponds to approximately 2.81 x 10^{21 electrons flowing through it. The calculation involves using the relationship between current, charge, and time, as well as the charge of an individual electron.
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