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hjlf
2 years ago
9

Why are the very first stars thought to have been much more massive than the sun? (a the clouds that made them were much more ma

ssive than todays star-forming clouds. (b the temperatures of the clouds that made them were higher because the clouds consisted entirely of hydrogen and helium. (c star-forming clouds were much denser early in time?
Physics
2 answers:
pashok25 [27]2 years ago
8 0
I think the correct answer would be that the temperature of the clouds that made the very first stars where thought to be higher since the clouds are made up of hydrogen and helium. Hope this answers the question. Have a nice day.
Juliette [100K]2 years ago
7 0
The main reason why the very first stars where thought to be much bigger than the sun because of the temperature of the clouds that these stars were made of were higher. The clouds that mostly form these stars consist of entirely hydrogen and helium. The answer for this would be option B.
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Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

3 0
2 years ago
It is known that heat is added to a gas in sealed container. The container is fitted with a moveable piston.
jasenka [17]

Answer:

d. Not enough information is given to answer this question.

Explanation:

From first law of thermodynamics

Q= W + ΔU

Q=Heat  ,W= Work , ΔU=Change in internal energy

If work done by the gas :

It means that W and Q both are positive

Q- W = ΔU

Ii Q > W ,then temperature of the gas will increase.

If  Q< W  ,Then temperature of the gas will decreases.

If work done on the gas:

Q positive but W will be negative

Q- W = ΔU

Q= W or Q>W  or Q< W ,then temperature of the gas will increase.

There are three cases because they did not give any information about the work.That is why option d is correct.

3 0
3 years ago
A 75-kg piano is hoisted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What
ipn [44]

Answer:

P = 14700 J

Explanation:

Given that,

Mass of a piano, m = 75 kg

It is delivered throughout the window of a 6th story apartment which is 20 m above the ground.

We need to find the potential energy of the piano. It is given by :

P = mgh

Putting all the values,

P = 75 kg × 9.8 m/s² × 20 m

P = 14700 J

So, the potential energy of the piano is 14700 J.

7 0
3 years ago
3.A bridge usually has expansion joints. They allow the bridge to become slightly longer when it
Dominik [7]

Answer:

a) When the sides of the joint are close together, the particles have more kinetic energy than they do when sides are farther apart.

Explanation:

5 0
3 years ago
I dont know how to do any of this so someone please help (the way it was solved has to be present for each thing) i will give br
Alexus [3.1K]
I cannot see all the questions, what is 18,19 and 21? (:
7 0
2 years ago
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