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kolbaska11 [484]
3 years ago
5

Which of the following is the most likely end for a star that is small to average in size? A. blue main sequence B. red supergia

nt C. supernova D. white dwarf
Physics
2 answers:
likoan [24]3 years ago
7 0
"White dwarf" is the one among the following choices given in the question that is <span>the most likely end for a star that is small to average in size. The correct option among all the options that are given in the question is the last option or option "D". I hope that this is the answer that has actually helped you.</span>
stiv31 [10]3 years ago
3 0

Answer: D. White Dwarf

Explanation:

A star has life-cycle similar to any living specie. A star takes birth, goes through main sequence and dies in the end. The lifetime and the end of the star depend on the mass it holds.

A massive star  turns into red super-giant and goes supernova and ends up as either a neutron star or a black-hole.

On the other hand, a small to average star like our Sun will become a red giant and throw extra mass before ending up as a white dwarf.

The color on the main sequence (blue, yellow or red) depends on the temperature and age of the star. A young hot star is blue main sequence star. On the other hand, an old low temperature star towards the end of its life is a red star.

Hence, end for a star that is small to average in size is most likely to be a D. white dwarf.


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HELP ME 20 POINTS!!!Assume that the average volume of an adult human body is one-tenth
Agata [3.3K]

Answer:

,Assume that the average volume of an adult human body is one-tenth

cubic meter (0.10 m) and that there are two billion (2.0 x 109)

adults in the world.

a. What would be the total volume of all the adults in the world?

b. Compute the length of one edge of a cubic container that has a

volume equal to the volume of all the adults in the world.

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3 years ago
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A bolt is dropped from a bridge under construction, falling 96 m to the valley below the bridge. (a) How much time does it take
irakobra [83]

Answer:

a)It takes the bolt 0.25 s to pass the last 11% of the fall.

b)When the bolt begins to fall the last 11% of the fall its velocity is -41.2 m/s.

c)The velocity of the bolt just before it reaches the ground is -43.6 m/s

Explanation:

Hi there!

a) Let´s calculate how much distance it is the last 11% of the fall:

96 m · 0.11 = 10.56 m

So, we have to find how much time it takes the bolt to pass from a height of 10.56 m to the ground.

First, let´s calculate how much time it takes the bolt to reach a height of 10.56 m. For that we can use this equation:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the bolt at a time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

If we consider the ground as the origin of the frame of reference, then h0 = 96 m. Since the bolt is dropped, the initial velocity is zero (v0 = 0). Then, the equation gets reduce to this:

h = h0 + 1/2 · g · t²

We have to find at which time h = 10.56 m.

10.56 m = 96 m - 1/2 · 9.8 m/s² · t²

Solving for t:

√(-2 · (10.56 m - 96 m) / 9.8 m/s²) = t

t = 4.2 s

Now that we have the time at which the bolt is located at 10.56 m above the ground, we can calculate the velocity of the bolt at that time.

The equation of velocity (v) of the bolt is the following:

v = v0 + g · t

at t = 4.2 s.

v = 0 - 9.8 m/s² · 4.2 s

v = -41.2 m/s

<u>When the bolt begins to fall the last 11% of the fall its velocity is -41.2 m/s.</u>

Now, we can calculate how much time it takes to fall the last 10.56 m.

The initial velocity of the bolt will be the velocity at h = 10.56 m. The initial height will be 10.56 m.

h = h0 + v0 · t + 1/2 · g · t²

We have to find the time at which h = 0 (the bolt hits the ground)

0 = 10.56 m - 41.2 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 0.25 s (the other solution of the quadratic equation is negative and thus discarded).

<u>It takes the bolt 0.25 s to pass the last 11% of the fall.</u>

Now, let´s calculate the velocity of the bolt when it reaches the ground:

v = v0 + g · t

v = -41.2 m/s - 9.8 m/s² · 0.25 s

v = -43.6 m/s

<u>The velocity of the bolt just before it reaches the bolt is -43.6 m/s</u>

6 0
3 years ago
What does it mean, if the specific gravity of an object is less than one?​
JulijaS [17]

Answer:

Specific gravity can be used to determine if an object will sink or float on water. ... If an object or liquid has a specific gravity greater than one, it will sink. If the specific gravity of an object or a liquid is less than one, it will float.

hope this helps, have a great day/night, and stay safe!

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Explain why mess extinctions occur and why it make sense to measure geologic time between mass extinctions
Nastasia [14]
Mass extinction occur from natural disasters, such as a n asteroid hitting earth or a volcano errupting and spread ash everywhere.

It makes sense to measure geologic time between mass extinctions because after each mass extinction, there is almost no life left and the few left have to repopulate, which may lead way to new mutations and new varieties of plants and animals.
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3 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
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