Ask question

Ask question

All categories

3 years ago

14

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

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

True or false? The difference between the state of matter (solid, liquid & gas) is their energy & bond

12345 [234]

True it’s true because in the book it said all that stuff

7 0

3 years ago

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

3 years ago

Which of these is a good example of Newton's First Law of inertia

Andru [333]

Answer:

one body movement to the side when a car makes a sharp turn

3 0

3 years ago

Read 2 more answers

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

anzhelika [568]

Answer:

The position is $P = 47.4 \ m$ relative to the base of the ocean

Explanation:

From the question we are told that

The angle made by the incline with the horizontal is $\theta = 24.0 ^o$

The constant acceleration is $a = 3.82 \ m/s^2$

The distance covered is $d = 60.0 \ m$

The height of the cliff is $h = 50 .0 \ m$

The velocity of the car is mathematically represented as

$v^2 = u^2 + 2ad$

The initial velocity of the car is u= 0

So

$v^2 = 2ad$

substituting values

$v^2 = 2 * 3.82 * 60$

$v = 21.4 \ m/s$

The vertical component of this velocity is

$v_v = -v * sin(\theta )$

substituting values

$v_v = -21.4 * sin(24.0)$

$v_v = -8.7 \ m/s$

The negative sign is because is moving in the negative direction of the y-axis

The horizontal component of this velocity is

$v_h = v * cos (\theta)$

$v_h = 21.4 * cos (24.0)$

$v_h = 19.5 \ m/s$

Now according to equation of motion we have

$h = v_v*t - \frac{1}{2} * g t^2$

substituting values

$50 = -8.7 t - \frac{1}{2} * 9.8 t^2$

$4.9t^2 +8.7t -50 = 0$

using quadratic equation we have that

$t_1 = 2.42\ s \ and\ t_2 = -4.20\ s$

given that time cannot be negative

$t = 2.42 \ s$

The car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

$P = v_h * t$

substituting values

$P = 19.5 * 2.43$

$P = 47.4 \ m$

5 0

3 years ago

What’s the equation for the alpha decay of plutonium-244.

kvv77 [185]

Answer:

Pu-244 and 94 alpha decay = U -240 and 92

8 0

3 years ago