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The velocity, density, and pressure of a certain fluid are related by [#permalink]
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18 Nov 2014, 08:35
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Re: The velocity, density, and pressure of a certain fluid are related by [#permalink]
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18 Nov 2014, 09:41
5v^2 for v=10, 5(10)^2 = 5(100) = 500. So now we have 500+P = c. Let's use c=1000. This means that P would be 500.
5v^2 for v=5, 5(25) = 125. So now we have 125 + P = c and since we used c=1000 above we have 125 + P =1000 or P=875.
875500 = 375 so we know that a decrease in velocity from 10m to 5m would raise the pressure by 375 pascals.
Answer C!



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Re: The velocity, density, and pressure of a certain fluid are related by [#permalink]
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18 Nov 2014, 11:45
Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution.Lets say V\(i\) as the initial velocity and V\(f\) as final velocity. similarly P\(i\) as the initial pressure and P\(f\) as final pressure. Given: V\(i\) = 10m/s V\(f\) = 5m/s therefore 5(10)^2 + P\(i\) = 5(5)^2 + P\(f\) P\(f\)  P\(i\) = 375. [C]
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Re: The velocity, density, and pressure of a certain fluid are related by [#permalink]
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19 Nov 2014, 00:19
as mentioned 5v^2+p=c p=c5v^2
initial v=10, p=c500........1 finial v=5, p=c125..........2
so difference of equations 1 & 2 will give value of pressure rise =375 pascal
Answer C



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Re: The velocity, density, and pressure of a certain fluid are related by [#permalink]
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19 Nov 2014, 01:32
Answer = C = 375 Initial \(P = C  5*10^2 = C  500\) After velocity decrease: \(P = C  5*5^2 = C  125\) Increase = C  125  (C  500) = 500  125 = 375 Answer = C
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Re: The velocity, density, and pressure of a certain fluid are related by [#permalink]
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19 Nov 2014, 07:53
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.
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The velocity, density, and pressure of a certain fluid are related by [#permalink]
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24 Jan 2018, 06:22
Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution. \({5(10)^2 + P }  {5(5)^2 + P } = 500  125 => 375\), Answer will be (C)
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