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katrin [286]
3 years ago
11

A star is in hydrostatic equilibrium when the outward push of pressure due to core burning is exactly in balance with the inward

pull of gravity. When the hydrogen in a star's core has been used up, burning ceases, and gravity and pressure are no longer in balance. This causes the star to undergo significant changes.
Which of the following evolutionary changes would bring a star back into hydrostatic equilibrium? Check all that apply:

a) A small increase in the star's internal pressure and temperature causes the star's outer layers to expand and cool.
b) A small decrease in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.
c) A small increase in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.
d) A small decrease in the star's internal pressure and temperature causes the star's outer layers to expand and cool.
Physics
1 answer:
viktelen [127]3 years ago
7 0

Answer:

a and b

Explanation:

Hydro static equilibrium holds a star steady and balanced. Whenever a star stops burning hydrogen in its center, there must be evolutionary improvements to maintain equilibrium for the star Of example, if a star's internal pressure and temperature fall, gravity will take over and force the star to contract and heat up, restoring stability. By contrast, if a star's internal pressure and temperature rises, the extra pressure causes the star to widen and cool, restoring balance.

so, according to above explanation options a and b both are true

a) A small increase in the star's internal pressure and temperature causes the star's outer layers to expand and cool.

b) A small decrease in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.

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A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
10points asap <br><br> A force of 30 N acts upon a 7 kg block. Calculate its acceleration.
nekit [7.7K]
Hello! Assuming that the only force acting on the mass is 30N...

Fnet = 30N
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I hope this helps!
5 0
3 years ago
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Answer:

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