Which of these is not an official SI unit of measure?

Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For

SpyIntel [72]

Answer:

$1.27\times 10^{12}\Omega/m$

Explanation:

We are given that

Diameter=d= $\mu m$

Thickness= $1\mu m$

Radius= $r=\frac{d}{2}=\frac{1}{2}\mu m=0.5\times 10^{-6} m$

Using $1\mu m=10^{-6} m$

Dielectric constant=8

Resistance = $R=2\times 10^5\Omega cm^2$

Internal specific resistance=r=100 ohm cm= $100\times \frac{1}{100}\Omega-m=1\Omega m$

Using 1 m=100 cm

Internal resistance per unit length= $\frac{r}{A}=\frac{1}{\pi r^2}=\frac{1}{3.14\times (0.5\times 10^{-6})^2}=1.27\times 10^{12}\Omega/m$

Using $\pi=3.14$

Internal resistance per unit length= $1.27\times 10^{12}\Omega/m$

8 0

3 years ago

You want to estimate the height of the Empire State Building. You start at its base and walk 15 m away. Then you approximate the

MatroZZZ [7]

Answer:

429 m

Explanation:

In this exercise you are suppose to simulate a right triangle, the catheti are the distance you walked and the height of the empire state building.

tan(88) = (height) /(distance you walked)

28.64 = height / 15

height = 429 m

8 0

3 years ago

A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca

Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, $f_o=30\ cm$

The focal length of the eyepiece lens of an astronomical telescope, $f_e=5\ mm=0.5\ cm$

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

$m=\dfrac{-f_o}{f_e}$

$m=\dfrac{-30}{0.5}$

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

7 0

3 years ago

What radiant energy is sensed by human skin as warmth

almond37 [142]

Infrared radiation <span>is sensed by human skin as warmth.</span>

4 0

3 years ago

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

$P.E=-\dfrac{GmM}{R+h}$

The kinetic energy at height above the surface of the earth

$K.E = 0$

The total energy at height above the surface of the earth

$E = K.E+P.E$

$E = -\dfrac{GmM}{R+h}$ ....(I)

The total energy at the surface of the earth

$E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$ ....(II)

We need to calculate the speed of the object just before it hits Earth

From equation (I) and (II)

$-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

Here, h = R

$-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

$v= \sqrt{\dfrac{GM}{R}}$

Hence, The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$ .

4 0

3 years ago