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kati45 [8]
3 years ago
10

As you learned in Part B, a non-burning helium core surrounded by a shell of hydrogen-burning gas characterizes the subgiant sta

ge of stellar evolution. As time goes on, the star continues to evolve, and eventually, it becomes a red giant. Rank the stages a star goes through as it evolves from a subgiant into a red giant, from latest to earliest.A) The shell of hydrogen surrounding the star's non-burning helium core ignites.B) Pressure in the star's core decreases.C) Pressure from the star's hydrogen-burning shell causes the non-burning envelope to expand.D) The star's non-burning helium core starts to contract and heat up.E) The star becomes a red giant.F) The surface of the star becomes brighter and cooler.
Physics
1 answer:
Gennadij [26K]3 years ago
6 0

Answer:

E- The star becomes a red giant (LATEST STAGE)

F- The surface of the star becomes brighter and cooler

C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand

A- The shell of hydrogen surrounding the star's nonburning helium core ignites.

D- The star's non burning helium core starts to contract and heat up

B- Pressure in the star's core decreases (EARLIEST STAGE)

(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.

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3 years ago
A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget
avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

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7 0
3 years ago
A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.
SashulF [63]

Answer:

Part a)

F = 0

Part b)

F = 0.25 N

Part c)

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

Explanation:

As we know that force on a current carrying wire is given as

\vec F = i(\vec L \times \vec B)

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

F = i(\vec L \times \vec B)

here we know that L and B is parallel to each other so

F = 0

Part b)

For 68.1 cm length wire we have

F = iLB sin\theta

here we know that

cos\theta = \frac{68.1}{166}

\theta = 65.8

so we have

F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8

F = 0.25 N

Part c)

For 151 cm length wire we have

F = iLB sin\phi

here we know that

cos\phi = \frac{151}{166}

\theta = 24.5

so we have

F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

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3 years ago
A car accelerates from 30 m/s to 50 m/s in 2 seconds. Calculate the cars acceleration
masha68 [24]

Answer:

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