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anastassius [24]

2 years ago

6

Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p

ressure must the water have as it moves slowly toward the nozzle that sprays it up into the air? Assume atmospheric pressure equal to 100,000. Pa and g=9.81 m/s2.

1 answer:

Brums [2.3K]2 years ago

6 0

Answer:

$P = 3.55 \times 10^5 Pa$

Explanation:

As we know that water from the fountain will raise to maximum height

$H = 26.0 m$

now by energy conservation we can say that initial speed of the water just after it moves out will be

$\frac{1}{2}mv^2 = mgH$

$v = \sqrt{2gH}$

$v = \sqrt{2(9.81)(26)}$

$v = 22.6 m/s$

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

$P = P_0 + \frac{1}{2}\rho v^2$

$P = 10^5 + \frac{1}{2}(1000)(22.6^2)$

$P = 3.55 \times 10^5 Pa$

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murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

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Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

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And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

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$F_y=W=mg=2.28 N$ (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

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