What body parts were scientists wanting to image that prompted the development of the CT scanner

What is meant in astronomy by the phrase "adaptive optics?

Sergeu [11.5K]

<h2>Answer: a.The mirrors and eyepiece of a large telescope are spring-loaded to allow them to return quickly to a known position. </h2>

Explanation:

Adaptive optics is a method used in several astronomical observatories to counteract in real time the effects of the Earth's atmosphere on the formation of astronomical images.

This is done through the insertion into the optical path of the telescope of sophisticated deformable mirrors supported by a set of computationally controlled actuators. Thus obtaining clear images despite the effects of atmospheric turbulence that cause the unwanted distortion.

It should be noted that with this technique it is also necessary to have a moderately bright reference star that is very close to the object to be observed and studied. However, it is not always possible to find such stars, so a powerful laser beam is used to point towards the Earth's upper atmosphere and create artificial stars.

7 0

3 years ago

Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

$Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s$

Que tengas un buen día!

6 0

3 years ago

when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv

sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0

3 years ago

A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction

Alja [10]

A) 1.55

The speed of light in a medium is given by:

$v=\frac{c}{n}$

where

$c=3\cdot 10^8 m/s$ is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

$v=1.94\cdot 10^8 m/s$

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

$n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55$

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda$ is the wavelenght in quartz

$\lambda_0$ is the wavelength in air

n is the refractive index

For the light in this problem, we have

$\lambda=355 nm\\n=1.55$

Therefore, we can re-arrange the equation to find $\lambda_0$ , the wavelength in air:

$\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm$

4 0

3 years ago

A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini

Margaret [11]

Answer:

$\mu_s=1.0205$

Explanation:

Given:

mass of solid disk, $m=2\ kg$

radius of disk, $r=2\ m$

force of push applied to disk, $F=20\ N$

distance of application of force from the center, $s=0.3\ m$

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

$\therefore F$

where:

$f_s$ = static frictional force

$\Rightarrow 20$

$\mu_s>1.0204$

7 0

2 years ago