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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:

You might be interested in
Which of the following maxims would apply to a person with a Type A personality?
lozanna [386]

The correct answer is C. Lost time is never found again

Explanation:

One of the most relevant personality theories is Type A and Type B personality theory that proposes two main types of personalities by grouping different personality traits. In the case of Type A personality, this refers to individuals that are very organized, impatient, and concerned with time and goals. On the opposite, Type B personality includes individuals that are more relaxed in different aspects.

According to this, one maxim that applies to Type A personality is "Lost time is never found again" because people with this personality are concerned about time and therefore, loss of time is considered highly negative by them. Also, due to their ambitions and concern with goals they want to avoid losing time as this is equivalent to work, money, goals, etc.

4 0
3 years ago
Explain the relationship between magnetic fields and magnetic force.
andriy [413]

Answer:

Magnetic field is the strength of magnetism created by a magnet, whereas the magnetic force is the force due to two magnetic objects. The concepts of magnetic field and magnetic force are widely used in fields such as classical mechanics, electromagnetic theory, field theory and various other applications.

Explanation:

3 0
3 years ago
What is the displacement from a starting position of (14.0, 3.0) m to a final position of (−3.0, −4.0) m?
dem82 [27]

Answer:

change in y = -7

change in x = -17

magnitude of displacement = sqrt(7^2+17^2)

tan of angle below -x axis = 7/17

because in third quadrant where x and y are negative

3 0
3 years ago
If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
3 years ago
The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter
Anuta_ua [19.1K]

The value of normal force as the slider passes point B is

  • 6 mg

The value of h when the normal force is zero

  • 3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N  = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

8 0
1 year ago
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