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3 years ago

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In a towns water system pressure gauges in still water at street level reads 150kpa. if a pipeline connected to the system break

s and shoots water straight up, how high above the street does the water shoot?

1 answer:

AnnZ [28]3 years ago

5 0

Answer:

The water shoots 15.31 m high above the street level.

Explanation:

The gauge pressure drives the motion of the water to whixhever height it will attain. The expression relating the gauge pressure to the height reached by the water, is

P = ρgh

P = Gauge Pressure = 150 kPa = 150,000 Pa

ρ = density of the fluid (water) = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = Height reached by the water = ?

150,000 = 1000 × 9.8 × h

h = (150000) ÷ 9800 = 15.306 = 15.31 m

Hope this Helps!!!

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Answer:

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Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

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Therefore,

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h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

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A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the ho

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You decide to roll a 0.11-kgkg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelen

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According to the de Broglie equation, matter can behave like waves, much like how light and radiation do, which are both waves and particles. A beam of electrons can be diffracted just like a beam of light, according to the equation. The de Broglie equation essentially clarifies the notion of matter having a wavelength.

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1 year ago

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

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The range of powers is $- 5 \ D \le P \le - 2.667\ D$

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From the question we are told that

The far point of the left eye is $n_f = 20 cm$

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From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

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To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

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converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

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To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

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$f_f = - \frac{20}{1} * \frac{1}{100}$

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Substituting values

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This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

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