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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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What size resistor would produce a current flow of 5 Amps with a battery voltage of 12.6 volts​
Debora [2.8K]

Answer:

resistance = 2.52 ohms

Explanation:

from the formula

V =IR

Voltage = (current)(resistance)

Resistance =

R=

R= 2.52 ohms

5 0
3 years ago
According to the place-time model of interaction, when people are in the same place, but at different times, this is an example
STALIN [3.7K]

Answer: Shift work communication.

Explanation:

Shift work communication is a type of communication that exist between shift workers, or people at different time zones.

Shift work consists of a three eight hours shift within 24 hours.

Keeping up with a shift work communication requires planning, time management and commitment. It also requires using various communication channels.

3 0
3 years ago
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0
3 years ago
2–25 Consider a medium in which the heat conduction equation is given in its simplest form as
barxatty [35]

Answer:

d) Is the thermal conductivity of the medium constant or variable.

Explanation:

As we know that

Heat equation with heat generation at unsteady state and with constant thermal conductivity given as

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}+\dfrac{\dot{q}_g}{K}=\dfrac{1}{\alpha }\dfrac{dT}{dt}

With out heat generation

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}

In 2 -D with out heat generation with constant thermal conductivity

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}

Given equation

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{a }\dfrac{dT}{dt}

So we can say that this is the case of  with out heat generation ,unsteady state and with constant thermal conductivity.

So the option d is correct.

d) Is the thermal conductivity of the medium constant or variable.

3 0
3 years ago
The cost to become a member of a fitness center is as follows: the membership fee per month is $50.00 the personal training sess
NikAS [45]

Answer:

#include <iostream>

using namespace std;

void information(){

cout<<"Welcome to the portal"<<endl;

cout<<"The membership fee per month is $50.00"<<endl;

cout<<"The personal training session fee is $30.00"<<endl;

cout<<"The senior citizens discount is 30%"<<endl;

cout<<"If the membership is bought and paid for 12 or more months, the discount is 15%;"<<endl;

cout<<"If more than five personal training sessions are bought and paid for, the discount on each session is 20%."<<endl;

}

void getInfo(bool &senior, int &months, int &personal){

cout<<"Are you senior citizen y/n ? : ";

char choice;

cin>>choice;

if(choice=='y' || choice=='Y'){

senior = true;

}else{

senior = false;

}

cout<<"Enter number of months for membership : ";

cin>>months;

cout<<"Enter number of personal training session : ";

cin>>personal;

}

double calcCost(bool senior, int months, int personal){

double cost=0;

double memberShipCost=months*50;

double trainingCost=personal*30;

if(personal>5){

trainingCost*=.80; //discount on personal training

}

if(months>=12){

memberShipCost*=.85; //discount on membership

}

cost = memberShipCost+trainingCost;

//discount for seniors

if(senior){

cost = cost*.70;

}

return cost;

}

int main() {

information();

cout<<"Select an option"<<endl;

char choice;

while(true){

cout<<endl<<endl;

cout<<"a. Calculate membership costs."<<endl;

cout<<"b. Quit program."<<endl;

cout<<"Enter your choice : ";

cin>>choice;

bool senior=true;

int months,personal;

if(choice=='a'){

getInfo(senior,months,personal);

double cost=calcCost(senior, months, personal);

cout<<"The calculated cost is $"<<cost<<endl;

}else if(choice=='b'){

break;

}

}

return 0;

}

Explanation:

8 0
4 years ago
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