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Ann [662]
2 years ago
13

Which cells or organs are considered to be part of both the immune and lymphatic systems? Select all that apply.

Physics
1 answer:
Leviafan [203]2 years ago
8 0

Answer:

lymph nodes

tonsils and adenoids

thymus

Explanation:

-Arteries are the blood vessels that take the blood that contains oxygen from the heart to the tissues and are part of the circulatory system.

-Lymph nodes are glands that take care of filtering the fluid that goes through the lympathic system and are also important for the functioning of the immune system.

-Capillaries are blood vessels that connect the veins and arteries and are part of the circulatory system.

-Tonsils and adenoids are located in the throat and they help protect the body from diseases and they are part of immune system and the lympathic system.

-Veins are the vessels that take the blood to the heart and they are part of the circulatory system.

-Thymus is an organ in which the T cells develop and they help protect the body against virus and bacteria and it is part of the immune and lympathic systems.

According to this, cells or organs that are considered to be part of both the immune and lymphatic systems are:

lymph nodes

tonsils and adenoids

thymus

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Alexxandr [17]

Answer:

a

  F  =326.7 \ N

b

  M  = 6

Explanation:

From the question we are told that

          The mass of the rock is  m_r  =  200 \ kg

          The  length of the small object from the rock is  d  =  2 \ m

          The  length of the small object from the branch l  =  12 \ m

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      W =  m_r *  g

substituting values

     W =   200 *  9.8

     W =   1960 \ N

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         \sum  M_f  =  F * cos \theta *  l  -  W cos\theta  *  d =  0

Here  \theta is very small so  cos\theta  *  l  =  l

                               and  cos\theta  *  d  =  d

Hence

       F *   l  -  W  * d =  0

=>    F  = \frac{W * d}{l}

substituting values

        F  = \frac{1960 *  2}{12}

       F  =326.7 \ N

The  mechanical advantage is mathematically evaluated as

          M  = \frac{W}{F}

substituting values

        M  = \frac{1960}{326.7}

       M  = 6

6 0
3 years ago
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

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Artist 52 [7]

Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

d = inside diameter = 8.1 m

P = internal pressure = 1.26 MPa

σ = 270 MPa

factor of safety = 2

Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?

The allow factor of safety:

\sigma _{a}  =\frac{\sigma }{factor-of-safety} =\frac{270}{2} =135MPa

The minimun wall thickness:

t=\frac{Pd}{4\sigma _{a} } =\frac{1.26*8.1}{4*135} =0.0189m

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Explanation:

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6 0
3 years ago
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