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1 year ago

Correctly label the following structures surrounding the small intestine. Stomach Appendix lleum Cecum Jejunum Ascending colon

1 answer:

7 0

The structures that are surrounding the small intestine in the image are: stomach, ascending colon, cecum and appendix.

<h3>What is the small intestine?</h3>

The small intestine will be part of the digestive system and it is the smallest part of the intestine, it will be after the stomach and before the ascending colon. It will be made up of three parts: duodenum, which is the one attached to the stomach, ileum, and jejunum.

Each of its parts is going to have certain specific functions related to digestion, absorption and have immunity as a function.

Among the characteristics of the small intestine, absorption stands out since its surface is created for better absorption, since it will have circular folds, intestinal villi and microvilli. These will increase the absorption surface, so it works better.

To learn more about small intestine visit: brainly.com/question/24180887

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A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels

Kryger [21]

Answer:

T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

$T = \dfrac{v^2.m}{l}$

$v = \dfrac{l}{s}$

$v = \dfrac{7.2}{0.74}$

v = 9.73 m/s

now, calculation of tension

$T = \dfrac{9.73^2\times 0.46}{7.2}$

T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0

3 years ago

Which nuclei is not radioactive? <br> A. Am-241<br> B. Mg-24<br> C. Pu-241<br> D. U-238

Helen [10]

Hello!

Which nuclei is NOT radioactive?

A) Am-241 B) Mg-24 C) Pu-241 D) U-238

Solving:

It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.

Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.

Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.

Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.

U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.

Answer:

B) Mg-24

_______________________

I Hope this helps, greetings ... Dexteright02! =)

6 0

3 years ago

Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul

Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

$v_1=\sqrt{2gh}$

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

$J=m(v_2-v_1)$

$J=2m\sqrt{2gh}$

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

7 0

3 years ago

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

3 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago