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ozzi
3 years ago
11

What is the maximum width for most vehicles?

Engineering
2 answers:
Vlad [161]3 years ago
6 0
8 feet hope this helps :)
Kazeer [188]3 years ago
5 0
The answer would be 8 feet dear!
You might be interested in
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 . For an applied str
erma4kov [3.2K]

Answer:

B. 2.3mm

Explanation:

The correct answer for the given question is B. 2.3mm. Some aircraft are made are fabricated from aluminium which has plane stain fracture toughness. In the given scenario the plane stain toughness is

Ktc = 40 [ MPa \sqrt{(m)} ]

When the Ktc increases the fracture will appear on the aircraft. In this case the maximum crack until the fracture failure is 2.5mm.

8 0
3 years ago
Consider film condensation on a vertical plate. Will the heat flux be higher at the top or at the bottom of the plate? Why?
STALIN [3.7K]

Answer:

During film condensation on a vertical plate, heat flux at the top will be higher since the thickness of the film at the top, and thus its thermal resistance, is lower.

Explanation:

https://www.docsity.com/pt/cengel-solution-heat-and-mass-transfer-2th-ed-heat-chap10-034/4868218/

https://arc.aiaa.org/doi/pdf/10.2514/1.43136

https://arxiv.org/ftp/arxiv/papers/1402/1402.5018.pdf

8 0
3 years ago
You do not need to remove the lead weights inside tires before recycling them.
almond37 [142]
There are NO “lead weights inside tires”!

Balancing of the entire wheel (rim plus tire ) is done by correctly fitting the tire to the rim to begin with (new tires have printed coloured marks identifying the heavy and light bits), and then by adding balancing weights to the metal rim, usually made of lead but not always.

If you’re recycling rims then yes, remove the weights and recycle those separately.
6 0
3 years ago
Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part
ra1l [238]

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

4 0
3 years ago
Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M
weqwewe [10]

Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

500 = \frac{P}{\frac{\pi}{4} 5^2}

solving for P

P = 9817.47 N

for Aluminium bar

150 = \frac{P}{\frac{\pi}{4} 10^2}

solving for P

P = 11790 N

aluminum bar carrying a higher load than steel bar

6 0
3 years ago
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