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dedylja [7]
2 years ago
7

Which component of the galaxy is shown in this image? Dust

Physics
1 answer:
Westkost [7]2 years ago
3 0

Answer:

stars

those are stars in the galaxy

You might be interested in
An acorn falls from a branch located 9.8 m above the ground. After 1 s of falling, the acorn's velocity will be 9.8 m/s downward
goldenfox [79]

Answer:

The acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

Explanation:

given information:

h =9.8

t =1 s

g = 9.8

the average speed

v = 1/2 gt²

   = 1/2 (9.8) (1)²

   = 4.8 m/s

the distance in 1s

h = v t

   = 4.8 (1)

   = 4.8 m

the acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

   =

8 0
3 years ago
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele
kirza4 [7]

Answer:

W = 462.5 keV

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

F = qE

F = (1.6 \times 10^{-19})(1.85 \times 10^6)

F = 2.96 \times 10^{-13} N

now the distance moved by the electron is given as

d = 0.25 m

so we have

W = F.d

W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)

W = 7.4 \times 10^{-14} J

now we have to convert it into keV units

so we have

1 keV = 1.6 \times 10^{-16} J

W = 462.5 keV

5 0
3 years ago
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
2 years ago
You place a solid cylinder of mass M on a ramp that is inclined at an angle β to the horizontal. The coefficient of static frict
zlopas [31]

Answer:

Explanation:

Let the frictional force required be f .

frictional force is responsible in creating rotational motion in the cylinder.

torque created by frictional force = f R

if angular acceleration be α

α = f R / I , I is moment of inertia of cylinder .

α = a / R , a is linear acceleration.

f R / I = a / R

a = f R² / I

linear acceleration a of cylinder down the slope

ma = mgsinθ - f  , ( f force is acting upwards and mgsinθ is acting downwards )

mf R² / I = mgsinθ -f

f ( m R² / I + 1) = mgsinθ

f = mgsinθ / ( m R² / I + 1)

= mgsinθ / ( m R² / mk² + 1) , k is radius of gyration of cylinder.

= mgsinθ / (  R² / k² + 1)

Putting the given values

f = Mgsinβ /(  R² / k² + 1)

for cylinder ,  R² / k² = 2

f =  Mgsinβ /3

6 0
3 years ago
Que fuerza será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s²
WARRIOR [948]

Answer:

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

Explanation:

La segunda ley de Newton, llamada ley fundamental o principio fundamental de la dinámica, plantea que un cuerpo se acelera si se le aplica una fuerza.

De esta manera, esta ley establece que las aceleraciones que experimenta un cuerpo son proporcionales a las fuerzas que recibe. Dicho de otra forma, la aceleración de un cuerpo es proporcional a la fuerza neta que se le aplica. Cuanto mayor es la fuerza que se le aplica a  un objeto con una masa dada, mayor será su aceleración.

La segunda Ley de Newton se expresa matemáticamente como:

F = m*a

Donde:

  • F es la fuerza neta. Se expresa en Newton (N)
  • m es la masa del cuerpo. Se expresa en kilogramos (Kg.).
  • a es la aceleración que adquiere el cuerpo. Se expresa en metros sobre segundo al cuadrado (m/s²).

En este caso:

  • m= 20 kg
  • a= 4 m/s²

Reemplazando:

F= 20 kg* 4 m/s²

Resolviendo:

F= 80 N

<u><em>La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.</em></u>

4 0
3 years ago
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