All categories

3 years ago

You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a

Physics

1 answer:

miv72 [106K]3 years ago

6 0

Answer:

Explanation:

Spiral Galaxies -

It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .

According to Hubble , these galaxies are classified as Sa , Sb , Sc .

Where ,

Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .

Sb - have a lesser bulge and the spiral is looser .

Sc - It has very weak bulge with the open spiral structure .

Hence , from the question ,

The given information is about the Sa .

You might be interested in

How can being scientifically literate help you in your own life?

stiv31 [10]

Answer: You can betterly understand what's around you and how it works

8 0

2 years ago

Two difference between lever and pulley

Annette [7]

Answer:

Differences can be :

1. Lever is onr of the six simple machines but pulley is a simple machine which has a wheel and axle .

2. Lever is of 3 types (first class , second class , third class ) but pulley is of only 2 types

I hope this helps you

7 0

3 years ago

A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o

dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?

we know $p = \frac{v^2}{R}$ .....................(i)

we need to find change in power

$\Delta P = \frac{\Delta (V^2)}{R}$

$\Delta P = \frac{2 V \Delta V}{R}$ ..............(ii)

from equations we get

$\frac{\Delta P}{P} = \frac{2 \Delta V}{V}$

$\frac{\Delta P}{P} = 2 \frac{110 -120}{120}$

$\frac{\Delta P}{P} = -2(\frac{10}{120})$

$\frac{\Delta P}{P} = - 0.16 %$

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller

7 0

3 years ago

Which method of heat transfer takes place when particles of matter vibrate and collide with each other in direct contact

Cerrena [4.2K]

Conduction, hope this helps! :)

6 0

3 years ago

Read 2 more answers

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago