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3 years ago

In stars mor massive than the sun, fusion continues until the core is almost all....

1 answer:

3 0

In stars more massive than the sun, the core temperature is hotter, which allows for fusion of more complex elements.

Most of the fusion occurs in the core.

In stars more massive than the sun, fusion continues through Deuterium, Carbon, and finally reaching iron/nickel.

Up to this point, the fusion reaction was endothermic, which means that the energy expended to produce the fusion reaction was exceeded by the energy produced in the reaction.

Fusion past iron is exothermic, and therefore the star will be able to survive by fusing elements heavier than iron.

After the core is almost entirely iron, the star is no longer in the Main Sequence.

So, fusion in stars more massive than the sun continue fusing until the core is almost entirely <em>iron</em>.

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For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.

Lelechka [254]

Answer:

B. 161.5 J

Explanation:

$n$ = Number of moles = 0.37

$\Delta T$ = Rise in the temperature of the oxygen gas = 15 K

$Q$ = heat added in order to raise the temperature

$c_{p}$ = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

$Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J$

6 0

3 years ago

The same element with different molecular structures are called?

jonny [76]

The answer is allotropes. Hope this helps. Have a great day.

6 0

3 years ago

Read 2 more answers

A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge

Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = $\frac{V}{R}$

where R is the resistance

R = $\frac{pL}{r^{2} \pi }$

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = $\frac{\pi(r)^{2}V }{pL}$

q = $\frac{\pi(r^{2} )Vt }{pL}$

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0

3 years ago

A toy balloon contains 0.75 L of helium at a pressure of 101 kPa. The balloon rises until the pressure on the balloon is 85 kPa.

Aleksandr [31]

<h3>The answer is 0.89 L</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{101000 \times 0.75}{85000} = \frac{75750}{85000} \\ = 0.89117647...$

We have the final answer as

Hope this helps you

4 0

3 years ago

you push a shopping cart with 2 packs of water bottles. you stop the cart and add two more packs. you push the cart again. does

HACTEHA [7]

Answer:

Yes

Explanation:

The cart will definitely behave differently because more weight is added to it.

In order to successfully push a cart, the applied force must exceed the force of the weight of the cart along with its constituents and the frictional force between the tyres and the floor.

<em>When more weights are added (the 2 packs of water bottles), it means that more force would be needed to overcome the opposing forces to the movement of the cart in order to successfully push the cart again.</em>

4 0

3 years ago