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Gennadij [26K]
1 year 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]1 year 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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Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr
Sedbober [7]

Answer:

(A) ratio of electric force to weight will be  23.469\times 10^{-10}

(b) Electric field will be E=4.26\times 10^{10}N/C

Explanation:

We have given mass of bee = 100 mg  = m=100\times 10^{-3}=0.1kg

Charge on bee q=23pC=23\times 10^{-12}C

Electric field E = 100 N/C

Weight of the bee W=mg=0.1\times 9.8=0.98N

Electric force on the bee F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N

So the ratio of electric force on the bee and weight is =\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}

(B) To hold the bee in air electric force must be equal to weight of bee

So mg=qE

0.1\times 9.8=23\times 10^{-12}E

E=4.26\times 10^{10}N/C

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HACTEHA [7]

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List the producers,consumers,and decomposers in food web
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Answer:sheep,wolf,ants

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false

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Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =
Ann [662]

Answer:

change in relative vorticity  0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C  per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g /  potential temperature ) × (potential vertical temperature) × exp^1/2    ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61  / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

8 0
3 years ago
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