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Ket [755]
3 years ago
13

Airbags will deploy in a head on collision but not in a collision that occurs from angle

Engineering
1 answer:
Aneli [31]3 years ago
8 0

Answer:  

Airbags will deploy in almost any angle.

Explanation:

Cars have sensors around them, so when the car gets hit, the sensors detect a crash and deploy the airbags to keep you safe.

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Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.​
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I believe it is cold forging?

sorry this answer was very confusing.
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Danny enjoys studying transportation networks and systems. Which field of engineering should he pursue?
andre [41]

Answer:

Civil Engineering.

Explanation:

Civil Engineering is a field of engineering study that deals with professional practising in designing and developing infrastructures such as transport system and networks, buildings, etc.

<u>Civil engineering is regarded as the second--oldest form of engineering studies in the field of engineering.</u>

<u>A person who studies civil engineering is able to design and develop major projects in transportations system such as road construction, bridges, airports, tunnels, etc.</u>

So, Danny, who is interested in studying transporation networks and system should opt for studying civil engineering.

8 0
3 years ago
How does the human body use phospholipids ?
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6 0
3 years ago
Read 2 more answers
A tee requires three studs and three cripples.<br> O TRUE<br> O FALSE
NARA [144]

Answer:

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3 0
3 years ago
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The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
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