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Mashcka [7]
3 years ago
13

How do the stars die?

Physics
1 answer:
mestny [16]3 years ago
5 0

A star’s death also depends on its mass. The most massive stars quickly exhaust their fuel supply and explode in core-collapse supernovae, some of the most energetic explosions in the universe. A supernova’s radiation can easily (if only briefly) outshine the rest of its host galaxy. The remnant stellar core will form a neutron star or a black hole, depending on how much mass remains. If the core contains between 1.44 and 3 solar masses, that mass will crush into a volume just 10 to 15 miles wide before a quantum mechanical effect known as neutron degeneracy pressure prevents total collapse. The exact upper limit on a neutron star mass isn’t known, but around 3 solar masses, not even neutron degeneracy pressure can combat gravity’s inward crush, and the core collapses to form a black hole.

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Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
17 points
const2013 [10]

The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>
  • MV = mv................. Equation 1

<h3>Where:</h3>
  • M = mass of the cannon
  • m = mass of the cannonball
  • V = velocity of the cannon
  • v = velocity of the cannonball

Make v the subject of the equation.

  • v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>
  • M = 500 kg
  • V = 3 m/s
  • m = 10 kg.

Substitute these values into equation 2.

  • v = (500×3)/10
  • v = 150 m/s.

 

Hence, The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0
2 years ago
A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output
Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

P=5\ V\times 1.97\ A

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

3 0
3 years ago
Which of the following creates an adhesive force that prevents separation of the parietal and visceral pleurae during ventilatio
Diano4ka-milaya [45]

Answer:

Negative intrapleural pressure is the correct answer

Explanation:

Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.

Air moves from a region of higher pressure to one of lower pressure.  Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.

Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.

3 0
3 years ago
What evidence do you that suggest water waves are transverse wave​
svp [43]

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

5 0
3 years ago
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