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Lana71 [14]
3 years ago
11

Heavier elements require higher temperatures to fuse. After stars run out of hydrogen in their cores, they leave the main sequen

ce, collapse, and eventually get hot enough to fuse the helium in their cores into carbon. As discussed in the previous chapter, the Sun will never get hot enough to fuse carbon.Which of these hypothetical situations would allow the temperature of the Sun's core to rise enough for carbon fusion to be possible?Choose one:A. Add mass to the Sun.B. Increase the radius of the Sun.C. Convert all of the Sun's hydrogen into carbon.D. Decrease the radius of the Sun.
Physics
1 answer:
Kruka [31]3 years ago
5 0

Answer:

A. Add mass to the Sun.

Explanation:

A. Add mass to the Sun.

Adding mass will make to take more time for the hydrogen to run out and hence, enough temperature will be developed to fuse helium atom into other heavier elements, and eventually get hot enough to fuse the helium in their cores into carbon.

The only hypothetical solution is that we need to add Mass to the Sun.

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Can a compound be separated by physical means?
Julli [10]

No,  it is not possible.
A compound is a substance or material constituting of two or more elements that have been chemically combined together to form a new, different substance

Any elements that have been joined together chemically can only be separated back into their constituent elements by chemical means because the bonds holding them together can only be broken using chemical means.

A good example is sodium chloride, table salt. Poisonous chlorine gas and toxic sodium metal react together whereby sodium loses one electron which chlorine readily accepts and in the process an ionic bond is formed between the two resulting in a totally new, harmless compound , sodium chloride.

Only through electrolysis can sodium chloride be separated back into sodium and chlorine gas. No physical means can be used to do that.

3 0
3 years ago
Which three quantities can be used to calculate acceleration?
ikadub [295]
The correct option will be
D. Time, initial velocity and final velocity
The Formula can be written as,
Acceleration=Final velocity-Initial Velocity/Time
5 0
3 years ago
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0
3 years ago
A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
jeyben [28]

Answer:

-4*10⁴ units.

Explanation:

As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must  still remain to be zero.

So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.

5 0
3 years ago
A car dropped from a height of 44 meters fall to a height of zero meters. How fast will the car be traveling as it hits the grou
FinnZ [79.3K]
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed

Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
Vf=29.3m/s
6 0
3 years ago
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