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Ad libitum [116K]
3 years ago
12

Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei

ght flow to be 500 lbf/s?
Physics
2 answers:
mars1129 [50]3 years ago
8 0

Explanation:

The given data is as follows.

   Fluid is water so, density \rho = 1000 kg/m^{3}

  Weight flow rate = 500 lbf/s = 2224.11 N/sec

  Cross-sectional area (A) = \pi \times \frac{30}{2} \times \frac{22}{2}

                                         = 0.05184 m^{2}

Hence, weight flow rate will be given as follows.

        w = \rho \times g \times A \times V

    2224.11 N/sec = \rho \times g \times A \times V

        V = \frac{2224.11}{1000 \times 9.81 \times 0.05184} m/s

            = 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

gizmo_the_mogwai [7]3 years ago
8 0

Answer:

Explanation:

length of major axis, 2a = 30 cm

a = 15 cm = 0.15 m

length of minor axis, 2b = 22 cm

b = 11 cm = 0.11 m

Weight flow, mg = 500 lbf/s = 2224.1 N/s

Area of duct, A = π ab = 3.14 x 0.15 x 0.11 = 0.0518 m²

Let v be the velocity

Volume per second = mass per second / density

Area x velocity = mass per second / density

0.0518 x v = 2224.1 / (9.8 x 1000)

v = 4.4 m/s

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Which of The following is the best example of water changing from a liquid to gas
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Answer:

There are no examples but this should be evaporation

Explanation:

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3 years ago
Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the hori
lys-0071 [83]

Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft

Explanation:

1. use the position (x) equation in parobolic movement to find the time (t)

565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°)  * t

t= 3.92 s

2. use the position (y) equation in parabolic movement to find de maximun heigth  the ball reaches at 565 ft from the home plate.

y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - \frac{32.2 ft/s^{2} *3.92 s^{2}  }{2}

y= 148.32 ft

3. finally add the 3 ft that exist between the home plate and the ball

148.32 ft + 3 ft = 151.32

6 0
3 years ago
If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is
expeople1 [14]

Answer:

We can use  2 g H = v2^2 - v1^2    or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph   we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

8 0
3 years ago
Calculate the heat flux (in W/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°C
ExtremeBDS [4]

Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

5 0
3 years ago
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