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podryga [215]
3 years ago
6

What kind of job does Malcolm have?

Engineering
2 answers:
Effectus [21]3 years ago
8 0
Huh? what do you mean?
velikii [3]3 years ago
4 0
His job was to examine product components for flaws and communicate their results to other team memebers
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Showing or hiding records in a database is called “filtering.”<br> True<br> False
agasfer [191]

Answer:

TRUE

Explanation:

4 0
2 years ago
Read 2 more answers
Which traits are common in all four career pathways of the Information Technology field? Check all that apply.
Komok [63]

Answer:

Accuracy and attention to detail, Problem solving and critical thinking skills, Knowledge of programming language .

Explanation:

It is a technological area in which a person learns how to develop computer hardware, including PCs, laptops, tablets, processing, networking, and other hardware parts. Another field of study in IT is Management Information Systems (MIS).

The IT industry's career paths can be categorized equally in the two primary field’s hardware and software areas  

In hardware, there is Production, maintenance, research and development, and strategic planning.

In software, there is manufacturing, development, programming, software testing, and maintenance and support under software.  

Computer operations, administration of databases, sales / marketing and data centre management are connected areas.

7 0
2 years ago
The pads are 200mm long, 150 mm wide and thickness equal to 12mm. 1- Determine the average shear strain in the rubber if the for
lord [1]

Answer:

a) 0.3

b) 3.6 mm

Explanation:

Given

Length of the pads, l = 200 mm = 0.2 m

Width of the pads, b = 150 mm = 0.15 m

Thickness of the pads, t = 12 mm = 0.012 m

Force on the rubber, P = 15 kN

Shear modulus on the rubber, G = 830 GPa

The average shear strain can be gotten by

τ(average) = (P/2) / bl

τ(average) = (15/2) / (0.15 * 0.2)

τ(average) = 7.5 / 0.03

τ(average) = 250 kPa

γ(average) = τ(average) / G

γ(average) = 250 kPa / 830 kPa

γ(average) = 0.3

horizontal displacement,

δ = γ(average) * t

δ = 0.3 * 12

δ = 3.6 mm

5 0
3 years ago
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
Urgent please help!<br> What are non-ferrous metal and ferrous metal?
m_a_m_a [10]
In metallurgy, non-ferrous metals are metals or alloys that do not contain iron in appreciable amounts. Generally more costly than ferrous metals, non-ferrous metals are used because of desirable properties such as low weight, higher conductivity, non-magnetic property or resistance to corrosion
8 0
2 years ago
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