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16 Sep 2014, 01:52
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C
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Difficulty:
25%
(medium)
Question Stats:
79% (01:37) correct
21%
(01:41)
wrong
based on 34
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The velocity, density, and pressure of a certain fluid are related by the equation \(5v^2 + P = c\), where \(v\) is the velocity in meters per second, \(P\) is the pressure in pascals, and \(c\) is a constant. If the velocity of this fluid decreases from 10 meters per second to 5 meters per second, by how many pascals does the pressure in the fluid rise?
A. 125
B. 250
C. 375
D. 500
E. 625
A. 125
B. 250
C. 375
D. 500
E. 625
16 Sep 2014, 01:52
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Official Solution:
The velocity, density, and pressure of a certain fluid are related by the equation \(5v^2 + P = c\), where \(v\) is the velocity in meters per second, \(P\) is the pressure in pascals, and \(c\) is a constant. If the velocity of this fluid decreases from 10 meters per second to 5 meters per second, by how many pascals does the pressure in the fluid rise?
A. 125
B. 250
C. 375
D. 500
E. 625
This problem requires the use of a special given equation, \(5v^2 + P = c\). We are told that the velocity decreases from 10 meters per second to 5 meters per second. The key to this problem is to recognize that we have a "before" situation and an "after" situation. In the "before" situation, we have a certain velocity and pressure; in the "after" situation, we have a different velocity and pressure. What remains constant between the two situations is the constant \(c\). Thus, we should set up the given equation for both situations.
To distinguish the situations, let's use subscripts on the variables. \(P_1\) and \(v_1\) will indicate "before," while \(P_2\) and \(v_2\) will indicate "after."
Before: \(5v_1^2 + P_1 = c\)
\(5(10)^2 + P_1 = c\)
\(500 + P_1 = c\)
After: \(5v_2^2 + P_2 = c\)
\(5(5)^2 + P_2 = c\)
\(125 + P_2 = c\)
Now, we cannot solve for \(c\), but we can set the left sides of these two equations equal to each other, because they are both equal to \(c\).
\(500 + P_1 = 125 + P_2\)
Again, we cannot solve for either pressure, but we do not need to. What we need to find is the increase in pressure - in other words, how much the pressure rises by. As an expression, the increase in pressure is simply \(P_2 - P_1\). Thus, we rearrange the equation to solve for this difference.
\(500 - 125 = P_2 - P_1\)
\(375 = P_2 - P_1\)
Answer: C
The velocity, density, and pressure of a certain fluid are related by the equation \(5v^2 + P = c\), where \(v\) is the velocity in meters per second, \(P\) is the pressure in pascals, and \(c\) is a constant. If the velocity of this fluid decreases from 10 meters per second to 5 meters per second, by how many pascals does the pressure in the fluid rise?
A. 125
B. 250
C. 375
D. 500
E. 625
This problem requires the use of a special given equation, \(5v^2 + P = c\). We are told that the velocity decreases from 10 meters per second to 5 meters per second. The key to this problem is to recognize that we have a "before" situation and an "after" situation. In the "before" situation, we have a certain velocity and pressure; in the "after" situation, we have a different velocity and pressure. What remains constant between the two situations is the constant \(c\). Thus, we should set up the given equation for both situations.
To distinguish the situations, let's use subscripts on the variables. \(P_1\) and \(v_1\) will indicate "before," while \(P_2\) and \(v_2\) will indicate "after."
Before: \(5v_1^2 + P_1 = c\)
\(5(10)^2 + P_1 = c\)
\(500 + P_1 = c\)
After: \(5v_2^2 + P_2 = c\)
\(5(5)^2 + P_2 = c\)
\(125 + P_2 = c\)
Now, we cannot solve for \(c\), but we can set the left sides of these two equations equal to each other, because they are both equal to \(c\).
\(500 + P_1 = 125 + P_2\)
Again, we cannot solve for either pressure, but we do not need to. What we need to find is the increase in pressure - in other words, how much the pressure rises by. As an expression, the increase in pressure is simply \(P_2 - P_1\). Thus, we rearrange the equation to solve for this difference.
\(500 - 125 = P_2 - P_1\)
\(375 = P_2 - P_1\)
Answer: C
09 Feb 2024, 03:59
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