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horrorfan
23 days ago
10

Interlocking stacked material is done by​ _____.

Engineering
2 answers:
iogann1982 [279]23 days ago
7 0

Sagot:

B. Ang paglalagay ng mga bagay sa tamang anggulo

Paliwanag:

Kailangan sundin ng mga manggagawa ang mga patakaran sa kaligtasan kapag humahawak ng mga stacked material. Ang isang materyal ay puwedeng bumagsak o mag-collapse na nagiging sanhi ng pinsala o pagkamatay. Ayon sa mga alituntunin ng OSHA, ang mga manggagawa ay dapat gawin ang mga sumusunod ngunit hindi limitado sa:

  • kapag gumagamit ng mga kamay, ang mga stack ay hindi dapat lumampas ng 16 talampakan ang taas at 20 talampakan ang taas kung gumagamit ng forklift
  • alisin ang mga paku mula sa ginamit na kahoy bago mag-stack
  • dapat matatag at self-stable ang mga stack
  • ang materyal na nakaimbak sa mga racks ay dapat nakaharap sa labas mula sa mga pangunahing daanan
Mrrafil [253]23 days ago
3 0

Sagot:

B

Paliwanag:

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A cylinder in space is of uniform temperature and dissipates 100 Watts. The cylinder diameter is 3" and its height is 12". Assum
Kisachek [217]

Answer:

Temperature T = 394.38 K

Explanation:

The full solution and detailed explanation regarding the above question and its specified conditions can be found below in the accompanying document. I trust my explanation will assist you in grasping this particular topic.

7 0
1 day ago
Write cout statements with stream manipulators that perform the following:
grin007 [219]

Answer:

A)cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B)cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C)cout<<fixed<<5.789E12;

D)cout<<left<<setw(7)<<67;

Explanation:

Stream Manipulators are special functions for use with the insertion (<<) and extraction (>>) operators on C++ stream objects, while the 'cout' statement outputs content to the standard output device in C++ programming.

setw: specifies the minimum width of the output field

setprecision: defines the number of decimal places for floating-point value formatting.

fixed: sets the format flag for floating-point streams.

left: left-aligns the output.

A) This statement shows the number 34.789 in a field that provides eight character spaces with two decimal precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B) Here, the number 7.0 is displayed within six spaces with three decimal precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C) This command prints 5.789e+12 in fixed-point format.  cout<<fixed<<5.789E12;

D) This statement left-aligns the number 67 across a field of six spaces. cout<<left<<setw(7)<<67;

7 0
23 days ago
estimate the area for a landfill for 12000 p producing waste for 10 y. assume that the national average is
alex41 [274]

Answer:

1.015 ha.

Explanation:

To calculate the landfill area required for 12,000 people producing waste over 10 years, follow these steps:[STEP ONE: Calculate the average solid waste generated per person per year (kg p^-1 ^y(kg/py)).

According to the problem, the average solid waste produced is 2.78 kg per person daily (kg/pd), hence converting to kg/py involves:

2.78 × 365 days = 1014.7 kg/py.

STEP TWO: Determine yearly volume of refuse per person.

Thus, volume = 1014.7 kg/py ÷ 500 kg/m^3 = 2.03 m^3 per person per year.

STEP THREE: Calculate total solid waste volume over 10 years for 12,000 individuals.

Total waste volume over 10 years = 10 × 12,000 × 2.03 = 243,600 m^3.

STEP FOUR: Find the required area for the landfill.

Note: The total height for the landfill should be 20 + 4 = 24m.

Thus, the area for the landfill = 243,600 m^3 / 24m = 10,150 m^2.

If 10,000 m^2 equals 1 ha, then 10,150 m^2 ÷ 10,000 m^2 = 1.015 ha.

(f). Ensure to expand the landfill area for enhancements.

4 0
21 hour ago
The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re
mote1985 [204]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, g = 9.81 m/s^2

mass flow rate, \dot{m} = 60.6 kg/s

flow density, \rho = 1000 kg/m^3

efficiency of the pump, \eta = 0.74

output gauge pressure, p_o = 344.75 kPa

input gauge pressure, p_i = 68.95 kPa

cross-sectional area of output pipe, A_o = 0.069 m^2

cross-sectional area of input pipe, A_i = 0.093 m^2

height of discharge, z_o = 1.22 m - 0.61 m = 0.61 m (evaluated at pump’s maximum height of 1.22 m)

input height, z_i = 0 m

hydraulic power of the pump,P =? kW

Initially, the volumetric flow (Q) needs to be determined

Q = \frac{\dot{m}}{\rho}

Q = \frac{60.6 kg/s}{1000 kg/m^3}

Q = 0.0606 m^3/s

Next, compute the velocity (v) for both input and output

v_o = \frac{Q}{A_o}

v_o = \frac{0.0606 m^3/s}{0.069 m^2}

v_o = 0.88 m/s

v_i = \frac{Q}{A_i}

v_i = \frac{0.0606 m^3/s}{0.093 m^2}

v_i = 0.65 m/s

Subsequently, the total head (H) can be calculated

H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}

H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}

H = 28.74m

Finally, the computation of pump power is done as follows

P = \frac{Q \, \rho \, g \, H}{\eta}

P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}

P = 23.09 kW

6 0
25 days ago
Effects of biological hazards are widespread. Select the answer options which describe potential effects of coming into contact
pantera1 [220]

Answer:

- Allergic Responses

- Events Posing Life Threats

Explanation:

Biological hazards can originate from a variety of sources such as bacteria, viruses, insects, plants, birds, animals, and humans. These can lead to numerous health issues, which may range from skin allergies and irritations to infections (like tuberculosis or AIDS) and even cancer.

7 0
27 days ago
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