An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Use the segment addition postulate RS + ST = RT to set up an equation for each problem.
Solve the equation to find the value of the variable ( a , x , or y ).
Substitute the value of the variable back into the expression for ST to find the length of segment ST .
The solutions are:
a = 4 , ST = 48
x = 11 , ST = 22
x = 5 , ST = 15
x = 2 , ST = 4
y = 4 , ST = 8
y = 2 , ST = 3
Explanation
Understanding the Problem We are given six different problems where point S lies between points R and T on a line. This means that the length of segment RS plus the length of segment ST equals the length of segment RT . In other words, RS + ST = RT . We will use this fact to solve each problem.
Solving Problem 28 For problem 28, we have RS = 7 a , ST = 12 a , and RT = 76 . Substituting these values into the equation RS + ST = RT , we get 7 a + 12 a = 76 . Combining like terms, we have 19 a = 76 . Dividing both sides by 19, we find a = 4 . Now we can find ST by substituting a = 4 into ST = 12 a , which gives ST = 12 ( 4 ) = 48 .
Solving Problem 29 For problem 29, we have RS = 12 , ST = 2 x , and RT = 34 . Substituting these values into the equation RS + ST = RT , we get 12 + 2 x = 34 . Subtracting 12 from both sides, we have 2 x = 22 . Dividing both sides by 2, we find x = 11 . Now we can find ST by substituting x = 11 into ST = 2 x , which gives ST = 2 ( 11 ) = 22 .
Solving Problem 30 For problem 30, we have RS = 2 x , ST = 3 x , and RT = 25 . Substituting these values into the equation RS + ST = RT , we get 2 x + 3 x = 25 . Combining like terms, we have 5 x = 25 . Dividing both sides by 5, we find x = 5 . Now we can find ST by substituting x = 5 into ST = 3 x , which gives ST = 3 ( 5 ) = 15 .
Solving Problem 31 For problem 31, we have RS = 16 , ST = 2 x , and RT = 5 x + 10 . Substituting these values into the equation RS + ST = RT , we get 16 + 2 x = 5 x + 10 . Subtracting 2 x from both sides, we have 16 = 3 x + 10 . Subtracting 10 from both sides, we have 6 = 3 x . Dividing both sides by 3, we find x = 2 . Now we can find ST by substituting x = 2 into ST = 2 x , which gives ST = 2 ( 2 ) = 4 .
Solving Problem 32 For problem 32, we have RS = 3 y + 1 , ST = 2 y , and RT = 21 . Substituting these values into the equation RS + ST = RT , we get ( 3 y + 1 ) + 2 y = 21 . Combining like terms, we have 5 y + 1 = 21 . Subtracting 1 from both sides, we have 5 y = 20 . Dividing both sides by 5, we find y = 4 . Now we can find ST by substituting y = 4 into ST = 2 y , which gives ST = 2 ( 4 ) = 8 .
Solving Problem 33 For problem 33, we have RS = 4 y − 1 , ST = 2 y − 1 , and RT = 5 y . Substituting these values into the equation RS + ST = RT , we get ( 4 y − 1 ) + ( 2 y − 1 ) = 5 y . Combining like terms, we have 6 y − 2 = 5 y . Subtracting 5 y from both sides, we have y − 2 = 0 . Adding 2 to both sides, we find y = 2 . Now we can find ST by substituting y = 2 into ST = 2 y − 1 , which gives ST = 2 ( 2 ) − 1 = 4 − 1 = 3 .
Final Answers In summary:
a = 4 , ST = 48
x = 11 , ST = 22
x = 5 , ST = 15
x = 2 , ST = 4
y = 4 , ST = 8
y = 2 , ST = 3
Examples
Imagine you're planning a road trip. You know the total distance you'll travel and the distance of the first leg of your journey. Using the segment addition postulate, you can easily calculate the distance of the remaining leg. This simple concept is fundamental in various real-world applications, from construction and engineering to everyday problem-solving involving distances and lengths.
The device delivers 450 Coulombs of charge while operating at 15.0 A for 30 seconds. This charge corresponds to roughly 2.81 × 1 0 21 electrons flowing through it. We calculated this by using the relationship between current, charge, and the charge of a single electron.
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