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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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A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North
ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km  

3 0
3 years ago
a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what
yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

i.e Force = Mass x Acceleration

In this case, Mass = ?

Force = 3.00 x 10^3 N = (3.00 x 1000N)

= 3000N

Uniform acceleration = 4.00m/s^2

Force = Mass x Acceleration

3000N = Mass x 4.00m/s^2

Mass = (3000N/4.00m/s^2)

Mass = 750Kg (The SI unit of mass is kilograms)

Thus, the mass of the car is 750Kg

4 0
3 years ago
El movimiento de los cuerpos y la formación de imágenes son ejemplos de esta clase de fenómeno ​
77julia77 [94]

Answer:

depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante

7 0
3 years ago
How are meteors and meteorites different?
konstantin123 [22]

A meteor is the flash of light that we see in the night sky when a small chunk of interplanetary debris burns up as it passes through our atmosphere. "Meteor" refers to the flash of light caused by the debris, not the debris itself.

If any part of a meteoroid survives the fall through the atmosphere and lands on Earth, it is called a meteorite.

5 0
3 years ago
Read 2 more answers
A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi
snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s            

Explanation:

It is given that,

Mass of the car, m_1=1.1\times 10^3\ kg

Mass of the truck, m_2=2.3\times 10^3\ kg

Initial velocity of the car, u_1=22\ m/s

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2    

v_2=15.7\ m/s

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0
3 years ago
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