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Shalnov [3]
3 years ago
12

A waste treatment pond is 50m long and 25m wide, and has an average depth of 2m.The density of the waste is 75.3 lbm/ft3. Calcul

ate the weight of the pond contents in lbf, using a single dimensional equation for your calculation.
Engineering
1 answer:
weeeeeb [17]3 years ago
6 0

Answer:

The weight of the pond contents is 6,643,233.22 lbf

Explanation:

Weight = density × volume × acceleration due to gravity

density = 75.3lbm/ft^3 = 75.3lbm/ft^3 × 0.45359kg/1lbm × (3.2808ft/1m)^3 = 1206.14kg/m^3

volume = length×width×depth = 50m×25m×2m = 2500m^3

acceleration due to gravity = 9.8m/s^2

Weight = 1206.14×2500×9.8 = 29,550,430N = 29,550,430N × 1lbf/4.4482N = 6,643,233.22lbf

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Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.
Liono4ka [1.6K]

Answer: Eye injury

Explanation: small material such as dust, dirt, and metal shards can harm your eyes with potential blindness or infection.

7 0
2 years ago
Which of the following is a variable expense for many adults?
sertanlavr [38]
But where are the options?
7 0
2 years ago
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the oscillation of rod oa about o is defined by the relation θ= (3/pi)*(sin*pi*t), where theta and t are expressed in radians an
Mashcka [7]

Answer:

(a) The velocity of the collar = 1.624er-15.54eo in/s

(b) the acceleration of the collar=-49.94er-9.74eo in/s²

(c)the acceleration of collative rod =-3.284er in/s²

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4 0
3 years ago
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 45 kPa when flying over the city. Calculate the loc
garri49 [273]

Answer:

1) The absolute pressure equals = 96.98 kPa

2) Absolute pressure in terms of column of mercury equals 727 mmHg.

Explanation:

using the equation of pressure statics we have

P(h)=P_{surface}-\rho gh...............(i)

Now since it is given that at 6400 meters pressure equals 45 kPa absolute hence from equation 'i' we obtain

P_{surface}=P(h)+\rho gh

Applying values we get

P_{surface}=45\times 10^{3}+0.828\times 9.81\times 6400

P_{surface}=96.98\times 10^{3}Pa=96.98kPa

Now the pressure in terms of head of mercury is given by

h_{Hg}=\frac{P}{\rho _{Hg}\times g}

Applying values we get

h=\frac{96.98\times 10^{3}}{13600\times 9.81}=0.727m=727mmHg

7 0
3 years ago
Water flows through a multisection pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s a
Alex_Xolod [135]

Answer:

b. 2.3 kPa.

Explanation:

This situation can be modelled by Bernoulli's Principle, as there are no energy interaction throughout the multisection pipe and current lines exists between both ends. Likewise, this system have no significant change in gravitational potential energy since it is placed horizontally on the ground and is described by the following model:

P_{1} + \rho \cdot \frac{v_{1}^{2}}{2} = P_{2} + \rho \cdot \frac{v_{2}^{2}}{2}

Where:

P_{1}, P_{2} - Pressures at the beginning and at the end of the current line, measured in kilopascals.

\rho - Water density, measured in kilograms per cubic meter.

v_{1}, v_{2} - Fluid velocity at the beginning and at the end of the current line, measured in meters per second.

Now, the pressure difference between these two points is:

P_{1} - P_{2} = \rho \cdot \frac{v_{2}^{2}-v_{1}^{2}}{2}

If \rho = 1000\,\frac{kg}{m^{3}}, v_{1} = 3\,\frac{m}{s} and v_{2} = 2.1\,\frac{m}{s}, then:

P_{1} - P_{2} = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \frac{\left(2.1\,\frac{m}{s} \right)^{2}-\left(3\,\frac{m}{s} \right)^{2}}{2}

P_{1} - P_{2} = -2295\,Pa

P_{1} - P_{2} = -2.295\,kPa (1 kPa is equivalent to 1000 Pa)

Hence, the right answer is B.

7 0
3 years ago
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