Each part of the mixture has the same appearance.

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di

vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi (\frac{D}{2})^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= \frac{d}{\pi}$

$r_e = \sqrt{\frac{d}{\pi} }$

replacing d = $\frac{\pi (\frac{D}{2})^2 }{10, 000}$ in the equation above; we have:

$r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }$

$r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}$

$r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

4 0

3 years ago

7.1 Project Guidelines 2021

Ede4ka [16]

I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and

3 0

2 years ago

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

slega [8]

Answer:

a) v_average = 11 m / s, b) t = 0.0627 s

, c) F = 7.37 10⁵ N

, d) F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

v² = v₀² - 2 a x

The fine speed zeroes them

a = v₀² / 2x

a = 22²/2 0.69

a = 350.72 m / s²

The average speed is

v_average = (v + v₀) / 2

v_average = (22 + 0) / 2

v_average = 11 m / s

b) The average time

v = v₀ - a t

t = v₀ / a

t = 22 / 350.72

t = 0.0627 s

c) The force can be found with Newton's second law

F = m a

F = 2100 350.72

F = 7.37 10⁵ N

.d) the ratio of this force to weight

F / W = 7.37 10⁵ / (2100 9.8)

F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

5 0

2 years ago

A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc

Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0

2 years ago

1–Find the components of each vector

navik [9.2K]

Answer:

ANSWERS ARE 100 % correct

Explanation:

trust me

7 0

2 years ago