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jasenka [17]
3 years ago
15

What engine does chrysler 300c have?​

Engineering
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer:

5.7 hemi v8

Explanation:

it has the 5.7 hemi v8 engine and the SRT comes with 6.1 version

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Serves as a protective barrier to prevent contact with engergized ("hot") parts<br> within the unit
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thanks hot hot

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If the driver gear has 5 teeth and the driven gear has 10 teeth,what is the gear ratio?​
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The on that has ten teeth
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How can statistical analysis of a dataset inform a design process
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2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11
ahrayia [7]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² =  1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² =  9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ =  (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )    

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵)  × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )    

0 - 0.6014925 + Rₓ =  0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R_y = m'( V₂sin(60°) - 0.5 )  

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R_y = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R_y = 0.092728( 0.305187 )

⇒ 1.5484 + R_y = 0.028299    

R_y = 0.028299 - 1.5484

R_y = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R_y² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

 

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Answer:

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