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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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Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

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a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

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