3 years ago

What happens to a low-mass star after helium flash?

Physics

1 answer:

ZanzabumX [31]3 years ago

4 0

The Luminosity Decreases.

You might be interested in

A child jumps on a trampoline. What causes the child to rise in the air?

nikdorinn [45]

Force causes him to go up

6 0

3 years ago

Read 2 more answers

Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine.

boyakko [2]

Answer:

Check explanation

Explanation:

In the formation of calcium fluoride we take calcium and fluorine.

in elemental form calcium exist in solid form and fluorine in gaseous form.

formation of compound takes place to complete their octet, in case of calcium need to remove two electron and need to add one elecron in fluorine to complete their octet so two electron will ransferred from calcium to two fluorine atom.

3 0

3 years ago

If you apply a force of 100 N to the level, how much force is applied to lift the crate?

Kaylis [27]

I believe that the answer is B. 133 N

4 0

3 years ago

Enny and Anne wanted to see who could throw a ball the hardest. They decided to each throw a ball against a wall as hard as they

mihalych1998 [28]

Answer:

wha t kind of variables

independent variables

dependent variables

controled variables

4 0

2 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago