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3 years ago

11

A motor boat is headed at a velocity of 25 kilometers/hour toward the north, while the velocity of the water current is 12 kilom

eters/hour to the west. What is the magnitude of the boat’s resultant velocity?

1 answer:

Tems11 [23]3 years ago

8 0

Explanation:

CON EL TEOREMA DE PITÁGORAS

<em>v</em> = $\sqrt{(12 km/h)^{2} + (25 km/h)^{2}}$ = 27.7 km/h

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Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

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$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

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So the equation becomes

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So we can write

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Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

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And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

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