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svp [43]
3 years ago
13

Which task best fits the role of a planning engineer?

Engineering
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

D

Explanation:

ensuring project end on time through carefully planning and organizing

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The waffle slab is: a) the two-way concrete joist framing system. b) a one-way floor and roof framing system. c) the one-way con
GREYUIT [131]

Answer:

a) the two-way concrete joist framing system

Explanation:

A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.

4 0
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you are planning to buy a new couch for your family room you measure the available space and conclude that the couch should be b
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5-6 feet is 60-72 inches and 8 feet is 96 inches
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3 years ago
Working with which of these systems requires a technician that has been certified in an EPA-approved course?
makvit [3.9K]

EPA Regulations provides a certified course for the technicians involved in the Air-conditioning system.

Answer: Option (b)

<u>Explanation:</u>

The EPA regulation has implemented an act called the "Clean Air Act" under the "section of 609".

This act provides some basic requirements for EPA Regulation such as follows;

  • Refrigerant: This unit must be approved by EPA Regulations before being implemented into the atmosphere.
  • Servicing: This system provides a certified course for technicians in service and also approve them with proper refrigerant equipment.
  • Reuse Refrigerants: The use of recycled refrigerants must be properly monitored before it comes in to serve.
6 0
3 years ago
How can I draw this image in 2D form
Ket [755]

Answer:

no it is not 2D

Explanation:

it is 3D

ok so follow these steps

- make hole

-make square

-make triangle

ok now your figure is ready

5 0
2 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
2 years ago
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