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Naily [24]
3 years ago
14

What does the term “retroperitoneal” mean in the terms of the location of the kidneys ?

Physics
1 answer:
lakkis [162]3 years ago
5 0

Answer:

"Retroperitoneal" refers to the back of the peritoneum, the membrane that lines the anatomical space in the abdominal cavity. Kidney stones may cause pain to the organs within the retroperitoneal space. A diagram of the aorta, a retroperitoneal structure.

Explanation:

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At what point is the northern hemisphere pointed farthest away from the sun?
inna [77]
<span>Well, It is the aphelion point, When the Earth is farthest away from the Sun, when the Northern Hemisphere is warm. the Earth is closest to the Sun, or at the perihelion, 2 weeks after the June Solstice, when the Northern Hemisphere is enjoying warm summer months. Well this kind of weather is very nice.</span>
7 0
3 years ago
An unknown substance from planet X has a density of 10 g/mL. It occupies a volume of 80 mL. What is the mass of this unknown sub
asambeis [7]

Answer:

800 mL

Explanation:

D*V=M

You pick out the numbers as well as what it is they represent from the word problem/explanation, then from there plug them in to the equations. Once you do that, you get your product and have the answer.

10*80= 800

5 0
3 years ago
iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig
Alika [10]

Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm  in air

8 0
2 years ago
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
Which would cause the largest increase in a mountain climber’s gravitational potential energy?
Gemiola [76]

Most likely climbing up the mountain

4 0
3 years ago
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