__5. The study of weather patterns can predict the trajectory and intensity of this

you read in an astronomy book that our milky way galaxy consists of about 200 billion stars. how do you suppose this number was

lapo4ka [179]

By utilizing a galaxy model that accurately explains all of the data that has been collected thus far.

<h3>Suppose it turns out that one in 1 million stars has a planet that at some point in its history is home to an advanced civilization. Then the total number of civilizations that have arisen in our galaxy would be closest to _____. Assume there are about 100 billion stars in the Milky Way galaxy.</h3>

It has been estimated that our galaxy has something close to 100,000 different civilizations.A galaxy is a collection of stars (or solar systems), gas, and dust that is located in a certain area of the universe.

It has been suggested that the Milky Way, the galaxy in which we reside, is replete with extinct civilizations.In conclusion, there are probably close to 100,000 civilizations in our galaxy as a whole.

To learn more about Milky Way Galaxy , visit :

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6 0

2 years ago

The ideal mechanical advantage of a machine reflects the increase or decrease in force there world be without friction, it is al

brilliants [131]

True: the ideal mechanical advantage of a machine is always greater than the actual mechanical advantage because all machines must overcome friction.

Explanation:

For a simple machine, it is possible to calculate two types of mechanical advantage:

1) The Ideal Mechanical Advantage (IMA) is given by

$IMA=\frac{d_e}{d_r}$

where

$d_r$ is the resistance arm

$d_e$ is the effort arm

The IMA gives the mechanical advantage of the machine if there are no friction forces acting on it, and if all the work in input is converted into work in output with no loss of energy

2) The Actual Mechanical Advantage (AMA) is given by

$AMA=\frac{L}{E}$

where

L is the load (the force in output)

E is the effort (the force in input)

The AMA gives the real mechanical advantage of the machine. For an ideal machine,

$AMA=IMA$

Because there is no loss of energy due to friction.

For a real machine instead,

$AMA$

because part of the input energy is converted into thermal energy and other forms of energy due to the presence of friction, so it is "wasted" energy.

Learn more about levers and machines:

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8 0

4 years ago

Salt water is denser than fresh water. a ship floats in both fresh water and salt water. compared to the fresh water, the volume

gavmur [86]

The ship floats in water due to the buoyancy Fb that is given by the equation:

Fb=ρgV, where ρ is the density of the liquid, g=9.81 m/s² is the acceleration of the force of gravity and V is volume of the displaced liquid.

The density of fresh water is ρ₁=1000 kg/m³.

The density of salt water is in average ρ₂=1025 kg/m³.

To compare the volumes of liquids that are displaced by the ship we can take the ratio of buoyancy of salt water Fb₂ and the buoyancy of fresh water Fb₁.

The gravity force of the ship Fg=mg, where m is the mass of the ship and g=9.81 m/s², is equal to the force of buoyancy Fb₁ and Fb₂ because the mass of the ship doesn't change:

Fg=Fb₁ and Fg=Fb₂. This means Fb₁=Fb₂.

Now we can write:

Fb₂/Fb₁=(ρ₂gV₂)/(ρ₁gV₁), since Fb₁=Fb₂, they cancel out:

1/1=1=(ρ₂gV₂)/(ρ₁gV₁), g also cancels out:

(ρ₂V₂)/(ρ₁V₁)=1, now we can input ρ₁=1000 kg/m³ and ρ₂=1025 kg/m³

(1025V₂)/(1000V₁)=1

1.025(V₂/V₁)=1

V₂/V₁=1/1.025=0.9756, we multiply by V₁

V₂=0.9756V₁

Volume of salt water V₂ displaced by the ship is smaller than the volume of sweet water V₁ because the force of buoyancy of salt water is greater than the force of fresh water because salt water is more dense than fresh water.

7 0

3 years ago

A thin plate 6 ft long and 3 ft wide is submerged and held stationary in a stream of water (T = 60°F) that has a velocity of 17

VashaNatasha [74]

A) In the case of the Boundary Thickness Layer we use the given formula,

$\delta = \frac{4.91x}{\sqrt{Re}}$

We know as well that,

Re = Número de Reynolds = $\frac{U*x}{\upsilon}$

Where,

U = velocity

$\upsilon$ = kinematic viscosity

For water, kinematic viscosity, $\upsilon = 1.21*10^{-5} ft^2 /s$

So, $500,000 = \frac{ 17x}{(1.21*10^{-5})}$

$x = 0.355 ft$

$d = \frac{4.91*0.355}{\sqrt {500000}}$

$d = 0.002465 ft = 0.029in$

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, $x = 0.355 ft$

C. Wall shear stress,

$\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }$

For water, dynamic viscosity, $\nu$ = 2.344*10^-5 lbf-s/ft^2

$\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}$

$\tau = 0.5605 lbf/ft^2$

4 0

3 years ago

You have two balls of equal size and smoothness, and you can ignore air resistance. One is heavy, the other is much lighter. You

motikmotik

Answer:

They will hit the ground at the same time.

Explanation:

By ignoring the opposing forces i.e. air resistant, both the heavy and light balls will fall with same acceleration due to gravity (g=9.8 m/s²) and g is independent of mass of the objects. Thus both will hit the ground at the same time.

4 0

3 years ago