Ask question

Ask question

All categories

3 years ago

13

The sparse area surrounding a spiral galaxy is called a

2 answers:

djyliett [7]3 years ago

3 0

<span>The sparse area surrounding a spiral galaxy is called a

Halo

</span>

Anit [1.1K]3 years ago

3 0

Answer: halo

Explanation: The sparse area surrounding a spiral galaxy is called a <em>halo</em>. The galactic halo extends beyond the main component. The galactic halo is composed by different elements of a galaxy. The stellar halo, the galactic corona and the dark matter halo.

You might be interested in

The d subshell can hold up to 10 electrons in an atom.<br> true or false

Lady_Fox [76]

Answer:

true

Explanation:

5 0

3 years ago

Read 2 more answers

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0

2 years ago

Explain biome in your own words.

EleoNora [17]

Hello,

A biome is a big habitat consisting of biotic and abiotic factors. there are 7 main biomes they are tundra.

taiga.

temperate deciduous forest.

scrub forest

grassland.

desert.

tropical rain forest.

temperate rain forest.

Hope this helps

Plz mark me as brainliest

3 0

3 years ago

Which one of the following formulas is used to find the voltage of a circuit?

Sedbober [7]

D , since Voltage is one joule per coulomb

8 0

2 years ago

Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

3 years ago