An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Identify the vertex of the quadratic from the table: ( 105 , 7 ) .
Write the vertex form of the quadratic equation: y = a ( x − 105 ) 2 + 7 .
Substitute the point ( 0 , 27 ) into the equation and solve for a : a ≈ 0.0018 .
The quadratic equation that models the situation correctly is: y = 0.0018 ( x − 105 ) 2 + 7 .
Explanation
Analyze the data We are given a table of x and y values and asked to find the quadratic equation that models the data correctly. The table is:
x 0 52.5 105 157.6 210 y 27 12 7 12 27 We can observe that the y -values are symmetric around x = 105 , which suggests that the vertex of the quadratic is at ( 105 , 7 ) .
Write the vertex form The general form of a quadratic equation in vertex form is given by:
y = a ( x − h ) 2 + k
where ( h , k ) is the vertex of the parabola. In our case, the vertex is ( 105 , 7 ) , so we have:
y = a ( x − 105 ) 2 + 7
Substitute a point Now we need to find the value of a . We can use one of the points from the table to solve for a . Let's use the point ( 0 , 27 ) . Substituting these values into the equation, we get:
27 = a ( 0 − 105 ) 2 + 7
Solve for a Now, we solve for a :
27 = a ( − 105 ) 2 + 7 27 − 7 = a ( 105 ) 2 20 = a ( 11025 ) a = 11025 20 = 2205 4 ≈ 0.001814
Find the equation Comparing the calculated value of a with the given options, we see that a ≈ 0.0018 . Therefore, the quadratic equation that models the situation correctly is:
y = 0.0018 ( x − 105 ) 2 + 7
Examples
Understanding quadratic equations can help model various real-world scenarios, such as the trajectory of a ball, the shape of a suspension bridge, or the profit margin of a business. In this case, the data could represent the height of a projectile at different times, and finding the quadratic equation allows us to predict its height at any given time. This is useful in fields like sports, engineering, and physics, where understanding parabolic motion is crucial for making accurate predictions and optimizing performance.
A current of 15.0 A flowing for 30 seconds results in a total charge of 450 Coulombs. This charge corresponds to approximately 2.81 x 10^21 electrons flowing through the device. Therefore, around 2.81 billion billion electrons pass through in this time period.
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