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3 years ago

17. Swing arm restraints are intended to prevent a vehicle from falling off a lift.

Engineering

1 answer:

Dimas [21]3 years ago

5 0

This is about usage of Swing arm restraints.

<em><u>B) False</u></em>

There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.

Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.

This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.

Read more at; brainly.com/question/17972874

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Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? (b) How

pogonyaev

Answer:

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) The kilograms of total ferrite and is 0.8311 kg

The kilograms of total cementite is 0.1689 kg

c) The kilograms of total cementite is 0.9343 kg

Explanation:

Given:

1 kg of austenite

1.15 wt% C

Cooled to below 727°C

Questions:

a) What is the proeutectoid phase?

b) How many kilograms each of total ferrite and cementite form, Wf = ?, Wc = ?

c) How many kilograms each of pearlite and the proeutectoid phase form, Wp = ?

d) Schematically sketch and label the resulting microstructure

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) To get the mass of the total ferrite form:

$W_{f} =\frac{C_{cementite}-C_{2} }{C_{cementite}-C_{1} }$

Here,

Ccementite = composition of cementite = 6.7 wt%

C₁ = composition of phase 1 = 0.022 wt%

C₂ = composition of alloy = 1.15 wt%

Substituting values:

$W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg$

To get the mass of the total cementite:

$W_{c} =\frac{C_{2}-C_{1}}{C_{cementite}-C_{1} } =\frac{1.15-0.022}{6.7-0.022} =0.1689kg$

c) The mass of pearlite:

$W_{p} =\frac{6.7-1.15}{5.94} =0.9343kg$

d) In the diagram you can see the different compositions (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)

5 0

3 years ago

Find the sum and product of each of these pairs of numbers. Express your answers as a base 3 expansion. Hint: We can just add an

Arlecino [84]

Answer:

$112_3 + 210_3 = 1022_3$

$112_3 \times 210_3 = 101220_3$

Explanation:

To find the sum of the two numbers, start summing from the right most numbers.

right most: 2 + 0 = 2

middle: 1 + 1 = 2

left: 1 + 2 = 3. Converting 3 to base 3, we have 1 r 0 = 10.

Therefore, $112_3 + 210_3 = 1022_3$ .

To find the product, simply do

$(0 \times 112) + (1 \times 112) + (2 \times 112)\\$ = $000 + 112 + 1001$ = $101220$ .

8 0

3 years ago

The dry unit weight of a soil sample is 14.8 kN/m3.

Jlenok [28]

Answer:

See attachment for completed question

Explanation:

Given that; Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

4 0

4 years ago

What is the angle of the input

olga55 [171]

Absolute positions — latitudes and longitudes
Relative positions — azimuths, bearings, and elevation angles
Spherical distances between point locations

3 0

3 years ago

What are the disadvantages of having a liquid cooled engine?

Feliz [49]

One notable disadvantage of liquid cooling over air cooling is that it is considerably costly to set up. Cooling fans are prevalent in the market, and this overabundance of supply means they are cheap. The components of a liquid cooling system can be expensive.

5 0

3 years ago