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julia-pushkina [17]
2 years ago
8

Why do giant stars become planetary nebulas while supergiant stars become supernovas when their nuclear fusion slows and is over

come by gravity?
Supergiant stars do not have enough mass to generate the gravity necessary to cause a planetary nebula.


Supergiant stars are too massive to have planets and thus do not have planetary nebulas.


Giant stars do not have enough mass to generate the gravity necessary to cause a supernova.


Supernovas require binary star systems to form and giant stars do not form binary star systems.
Engineering
1 answer:
sashaice [31]2 years ago
3 0

The reason why giant stars become planetary nebulas is  Supergiant stars do not have enough mass to generate the gravity necessary to cause a planetary nebula.

<h3>Why do giant stars become planetary nebulae?</h3>

A planetary nebula is known to be formed or created by a dying star. A red giant is known to be unstable and thus emit pulses of gas that is said to form a sphere around the dying star and thus they are said to  be ionized by the ultraviolet radiation that the star is known to releases.

Learn more about  giant stars from

brainly.com/question/27111741

#SPJ1

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The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect
Yakvenalex [24]

Answer:

false

Explanation:

5 0
3 years ago
Read 2 more answers
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

6 0
3 years ago
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
3 years ago
5. Assume that you and your best friend ench have $1000 to invest. You invest your money
Bezzdna [24]

Correct question reads;

Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:

(a) You have $10 more than she does

(b) You have $100 more than she does

(c) You both have the same amount of money

(d) She has $10 more than you do

<u>Answer:</u>

<u>(d) She has $10 more than you do</u>

<u>Explanation</u>:

Using the compound interest formula

A= P [ (1-i)^n-1

Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.

<u>My compound interest is:</u>

= 1000 [ (1-0.1)^1-1

= $1000

$1,000 + $1,000 invested= $2,000 total amount received.

<u>My friend's simple interest is;</u>

To determine the total amount accrued we use the formula:

P(1 + rt) Where:

P = Invested Amount (1000)

I = Interest Amount (10,000)

r = Rate of Interest per year (10% or 0.2)

t = Time Period (1 )

= 1000 (1 + rt)

= 1000 (1 + 0.1x1)

= $1100 + $1000 invested = $2100 total amount received.

Therefore, we observe that she (my friend) has $100 more than I do.

5 0
3 years ago
A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp
Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature  = - 35 degree C

Maximum expansion can be determined as

\Delta L = L \alpha (T_{max} - T_{min})

where , \alpha is expansion coefficient = 12\times 10^{-6} degree C

SO, \Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))

\Delta L = 0.45 m = 450 mm

number of minimum expansion joints is calculated as

n = \frac{450}{20} = 22.5

b) length of each bridge

Length = \frac{500}{22.5} = 22.22 m

8 0
3 years ago
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