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kupik [55]
2 years ago
4

Type-Ia supernovae are probably thermonuclear explosions of accreting white dwarfs that have approached or reached the Chandrase

khar limit. a. Use the virial theorem to obtain an expression for the mean pressure inside a white dwarf of mass M and radius R. b. Use the result of (a) to estimate, to an order of magnitude, the speed of sound, vs = p dP/drho ∼ p P/rho, inside a white dwarf. In an accreting white dwarf with a carbon core that has reached nuclear ignition temperature, a nuclear burning "flame" encompasses the star at the sound velocity or faster. Within how much time, in seconds, does the flame traverse the radius of the white dwarf, assuming R = 104 km, M = 1.4M¯? Note that this soundcrossing timescale is ∼ (Grho) −1/2 , which is also the free-fall timescale (Eq. 3.15.)

Physics
2 answers:
blondinia [14]2 years ago
8 0

Answer:

a. see attachment

b. 2.5 seconds

Explanation:

Chandrasekhar limit:

The Chandrasekhar limit can be seen as the maximum mass of a stable white dwarf star. The currently accepted value of the Chandrasekhar limit is about 1.4 M (2.765×1030 kg).

Using the information above, kindly check attachment for details of the solution.

nordsb [41]2 years ago
8 0

Answer:

(a) P = (3/20)(πGnM²/R⁴)

(b) t = 2.5 seconds

Explanation:

See the attached file for rxplanation

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You are using a rope to lift a 14.5 kg crate of fruit. Initially you are lifting the crate at 0.500 m/s. You then increase the t
lina2011 [118]

Answer:

W = 172.5 J

Explanation:

given,                                    

mass of the fruit crate = 14.5 kg

initial velocity to lift = 0.500 m/s

increase in the tension = 150 N

lift of crate = 1.15 m                  

work done by the tension = ?        

work done  = force x displacement

W = F s cos θ                                

θ = 0°                                    

W = F s x cos 0                                  

W = 150 x 1.15 x 1                

W = 172.5 J                                      

Work done on the crate by the tension force = W = 172.5 J

5 0
2 years ago
Which of the following selections completes the following nuclear reaction?
Monica [59]

Answer:

n (a neutron)

Explanation:

For a chemical element:

- The lower subscript indicates the atomic number (the number of protons)

- The upper subscript indicates the mass number (the sum of protons and neutrons in the nucleus)

In the reaction described in the problem, we see that a gamma photon hits a nucleus of Calcium-40, which has

Z = 20 (20 protons)

A = 40 (40 protons+neutrons)

Which means that the number of neutrons is n =  A - Z = 40 - 20 = 20

After the reaction, we have a nucleus of Calcium-39, which has

Z = 20 (20 protons)

A = 39 (39 protons+neutrons)

Which means that the number of neutrons is n =  A - Z = 40 - 39 = 19

So, the nucleus has lost 1 neutron, which is the particle missing in the reaction.

3 0
2 years ago
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
Calculate the pressure, in atmospheres, exerted by each of the following:
gregori [183]

Answer:

a) 14.2 atm

b) 4.46 atm

c) 1.06 atm

Explanation:

For an ideal gas,

PV = nRT

P = pressure of the gas

V = volume occupied by the gas

n = number of moles of the gas

R = molar gas constant = 0.08206 L.atm/mol.K

T = temperature of the gas in Kelvin

a) For HF,

P =?, V = 2.5L, n = 1.35 moles, T = 320K

P = 1.35 × 0.08206 × 320/2.5

P = 14.2 atm

b) For NO₂

P =?, V = 4.75L, n = 0.86 moles, T = 300K

P = 0.86 × 0.08206 × 300/4.75

P = 4.46 atm

c) For CO₂

P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K

P = 2.15 × 0.08206 × 330/55

P = 1.06 atm

4 0
2 years ago
The hammer throw was one of the earliest Olympic events. In this event, a heavy ball attached to a chain is swung several times
Aleonysh [2.5K]

Answer:

Given that

T= 0.43 s

Radius of the ball path's , r=2.1 m

a)

We know that

f= 1/T

Here f= frequency

      T= Time period

Now by putting the values

f= 1/T

T= 0.43 s

f= 1/0.43

f=2.32 Hz

b)

We know that

V= ω r

ω = 2 π f

ω=Angular speed

V= Linear speed

ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s

V= ω r= 14.60 x 2.1 = 30.66 m/s

c)

Acceleration ,a

a =ω ² r

a= 14.6 ² x 2.1 = 447.63 m/s²

We know that g = 10 m/s²

So

a= a/g= 447.63/10 = 44.7 g m/s²

a= 44.7 g m/s²

7 0
2 years ago
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