All categories

3 years ago

Who’s has bigger cross sectional area capillaries or aorta?

Physics

1 answer:

Natalka [10]3 years ago

8 0

Answer:

Even though the cross-sectional area of each capillary is extremely small compared to that of the large aorta, the total cross-sectional area of all the capillaries added together is about 1,300 times greater than the cross-sectional area of the aorta because there are so many capillaries

Explanation:

You might be interested in

In what way does Isaac represent us?

Andrews [41]

Answer:

Derived from the Hebrew יִצְחָק (Yitzhak), the name Isaac means “one who laughs” or “one who rejoices.” In the Old Testament of the Bible, Isaac is the firstborn son of Abraham. He is one of the three biblical patriarchs revered by Jews, Christians, and Muslims. ... Gender: Isaac is traditionally a masculine name.

7 0

2 years ago

What area of science compares and describes quantities and movements of matter and energy?

qaws [65]

Answer:

the answer is physics

4 0

3 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

7 0

3 years ago

Determine la inercia rotacional de una varilla de 4 m de largo y 2 Kg de mesa si su eje de rotación esta situado a la mitad de l

Maslowich

Answer:

I = 2.667 kg m²

Explanation:

The moment of inertia of a body can be calculated by the expression

I = ∫ L² dm

For high symmetry bodies the expressions of the moment of inertia are tabulated, for a rod with its axis of rotation at its midpoint it is

I = $\frac{1}{12}$ m L²

let's calculate

I = $\frac{1}{12}$ 2 4²

I = 2.667 kg m²

8 0

3 years ago

When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the

Masja [62]

Answer:

The smallest angular velocity is 4.42 rad/s and the largest angular velocity is 13.15 rad/s for the toast to hit and then topple to be butter-side down

Explanation:

62 cm = 0.62 m

Knowing that the vertical distance traveled is s = 0.62 m and gravitational acceleration is g = 9.81 m/s2, we can calculate the time of falling

$s = gt^2/2$

$0.62 = 9.81t^2/2$

$t^2 = 2*0.62/9.81 = 0.13$

$t = \sqrt{0.13} = 0.36 s$

If there's only 0.36s to rotate, and for the toast to hit and then topple to be butter-side down, then the angle of rotation must be between 90 and 270 degrees (Suppose that it starts with butter up and it rotates less than 1 revolution, or 360 degrees):

In other words (and units):

$\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$ rad

As $\omega = \theta / t$ :

$\frac{\pi}{2} / t \leq \theta / t \leq \frac{3\pi}{2} / t$ rad/s

$\frac{\pi}{2*0.36} \leq \omega \leq \frac{3\pi}{2*0.36}$ rad/s

$4.42 \leq \omega \leq 13.25$ rad/s

So the smallest angular velocity is 4.42 rad/s and the largest angular velocity is 13.15 rad/s for the toast to hit and then topple to be butter-side down

4 0

3 years ago