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# The velocity, density, and pressure of a certain fluid are related by

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Math Expert
Joined: 02 Sep 2009
Posts: 49320
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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5% (low)

Question Stats:

90% (01:10) correct 10% (00:28) wrong based on 62 sessions

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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.

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Posts: 98
Re: The velocity, density, and pressure of a certain fluid are related by  [#permalink]

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18 Nov 2014, 09:41
1
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.

875-500 = 375 so we know that a decrease in velocity from 10m to 5m would raise the pressure by 375 pascals.

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Joined: 25 May 2014
Posts: 46
Re: The velocity, density, and pressure of a certain fluid are related by  [#permalink]

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18 Nov 2014, 11:45
1
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
1
as mentioned 5v^2+p=c
p=c-5v^2

initial v=10, p=c-500........1
finial v=5, p=c-125..........2

so difference of equations 1 & 2 will give value of pressure rise =375 pascal

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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
2

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

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Posts: 49320
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.
$$500 - 125 = P_2 - P_1$$
$$375 = P_2 - P_1$$

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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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