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Gennadij [26K]
2 years ago
7

Asymmetric dimethylarginine, endocan, pentraxin 3, serum amyloid A, soluble urokinase plasminogen activator receptor, total oxid

ant status and total antioxidant status, galectin-3
Physics
1 answer:
stealth61 [152]2 years ago
6 0

relationship between these inflammatory markers and infection.

<h3>What is Inflammatory markers ?</h3>

Recognised inflammatory biomarkers play important roles in the management of patients with for example, diagnosis, follow-up, assessment of treatment response and risk stratification

  • Asymmetric dimethylarginine, endocan, pentraxin 3, serum amyloid A, soluble urokinase plasminogen activator receptor, total oxidant status and total antioxidant status, galectin-3 were considered among the emerging inflammatory markers
  • Well-known inflammatory markers included complete blood count, C-reactive protein, albumin, cytokines and erythrocyte sedimentation rate.

Learn more about Inflammatory markers here:

brainly.com/question/8882360

#SPJ4

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Using what you know about the periodic table, choose all of the correct answers,
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A thin, light wire 75.1 cm long having a circular cross section 0.555 mm in diameter has a 25.4 kg weight attached to it, causin
seraphim [82]

Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

(a) Stress = Force/Area.

Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²

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Read 2 more answers
A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally
leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

3 0
3 years ago
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