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3 years ago

5x - 1 1 - 2x3511.152<8+2

Physics

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

x≥0

Explanation:

Let's find the critical points of the inequality.

−2x3511.152+5x−11=10

−2x3511.152+5x−11+−5x=10+−5x(Add -5x to both sides)

−2x3511.152−11=−5x+10

−2x3511.152−11+11=−5x+10+11(Add 11 to both sides)

−2x3511.152=−5x+21

−2x3511.152

−2=−5x+21/−2

(Divide both sides by -2)

x3511.152=5/2x+−21/2

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Answer:

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Explanation:

From the question we are told that

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The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

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Here $v_2$ is the new velocity

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

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=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

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2 years ago

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Answer:

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3 years ago

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Answer:

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3 years ago

5x - 1 1 - 2x3511.152<8+2​

5x - 1 1 - 2x3511.152<8+2