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3 years ago

What are 3 types of cells that helper T cells produce upon being an alarmed of an antigen? Thanks for your help!! :)

1 answer:

5 0

Whenever these are an alarming due to a foreign antigen, the Helper T cells, Memory T cells, and Cytotoxic T cells are being produced. Helper T cells to recognize the specific foreign antigen that is entering the body, Memory T cells to remember the specific foreign antigen that is attacking, and Cytotoxic T cells that are for the destruction of the foreign cells and produce cytokines to attract macrophages

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every object that has mass attracts every other object with a gravitational force. it has been proven that the size of the gravi

erastova [34]

Answer;

- This statement about matter and its behavior is best classified as a Law.

-It is the Law of Universal Gravitation.

Explanation;

-The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation.This attractive force always points inward, from one point to the other.

-The Law applies to all objects with masses, big or small. Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric.

3 0

3 years ago

Read 2 more answers

Which term describes the point when two objects reach the same temperature and there is no longer a transfer of energy through h

yaroslaw [1]

The term you need to know is equilibrium. Technically it means that heat gained = heat lost. Normally in beginning chemistry classes the evidence for this condition is a stable temperature.

5 0

3 years ago

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

3 years ago

If a rock is thrown vertically upward from the surface of Mars with velocity of 20 m/s, its height (in meters) after t seconds i

8_murik_8 [283]

The velocity of the rock after 2s is 12.56 m/s.

The velocity of the rock at 25 m upwards is 14.68 m/s.

The velocity of the rock at 25 m downwards is -15.97 m/s.

The given parameters;

velocity of the rock, u = 20 m/s
height of the rock, h = 20t - 1.8t²

The velocity of the rock after 2s is calculated as follows;

$v = \frac{dh}{dt}= 20 -2(1.86)t \\\\v = 20 - 3.72t\\\\at \ t= 2 \ s \\\\v = 20 - 3.72(2) = 12.56 \ m/s$

The velocity of the rock when the height is 25 m:

h = 20t - 1.8t²

25 = 20t - 1.8t²

1.8t² - 20t + 25 = 0

solve for the time of motion "t" using quadratic formula

a = 1.8, b = -20, c = 25

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-20) \ \ +/- \ \ \sqrt{(-20)^2 - 4(1.8\times 25)} }{2(1.8)}\\\\t = 9.67s, \ \ or \ 1.43 \ s$

The value of t that will give 25 m upwards using the motion model is 1.43 s;

h = 20(1.43) - 1.86(1.43)² = 25 m

The velocity of the rock at 25 m upwards (t = 1.43 s) is calculated as follows;

$v = \frac{dh}{dt} = 20 -3.72t\\\\v = 20 - 3.72(1.43) \\\\v = 14.68 \ m/s$

The velocity of the rock at 25 m downwards (t = 9.67 s) is calculated as;

$v = \frac{dh}{dt} = 20 - 3.72t\\\\v = 20 - 3.72(9.67)\\\\v = -15.97 \ m/s$

Learn more here:brainly.com/question/25233377

5 0

3 years ago

It is raining outside, so a student lifts their textbook vertically 1.2 meters above

Nadusha1986 [10]

Answer:

Work done is 50.4J.

Explanation:

The work $W$ done by the student in lifting the book 1.2 meters vertically is

$W =42N*1.2m$

$\boxed{W = 50.4J}$

When the student begins to walk horizontally, the net force on the book is zero (gravity is pulling it down, the student is lifting it up). Thus, although the student is displacing the book horizontally, but because no force is exerted on it, the work done is zero.

3 0

4 years ago