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3 years ago

The classical pathway for complement activation is initiated by

1 answer:

maks197457 [2]3 years ago

4 0

Answer: The classical complement pathway for complement activation is initiated by antigen-antibody complexes with the antibody isotypes IgG and IgM.

Explanation: The classical complement pathway typically requires antigen-antibody complexes (immune complexes) for activation (specific immune response), whereas the alternative pathway can be activated by C3 hydrolysis, foreign material, pathogens, or damaged cells.

After activation, a series of proteins are recruited to generate C3 convertase, which cleaves the C3 protein. The C3b component of the cleaved C3 binds to C3 convertase to generate C5 convertase, which cleaves the C5 protein. The cleaved products attract phagocytes to the site of infection and tags target cells for elimination by phagocytosis. In addition, the C5 convertase initiates the terminal phase of the complement system, leading to make appear the membrane attack complex. The membrane attack complex creates a pore on the target cell's membrane, inducing cell lysis and death.

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Describe how excavations are done.

Mazyrski [523]

Answer:

It includes earthwork, trenching, wall shafts, tunneling and underground

Explanation:

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3 years ago

What is a learned behavior of a puppy

Ludmilka [50]

Its when the animal repeats what it is taught similar to a child you never really teach them to lie it just happens because they have seen other people do it

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3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i

evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 $\frac{m}{s}$

To find:

the total time it is in the air = ?

Formula used:

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = $\frac{0-8}{-9.8}$

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

4 0

3 years ago

A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample

Musya8 [376]

Answer:

remain 1s60

Explanation:

I took away sample

7 0

2 years ago