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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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HELP
Sergeu [11.5K]

Answer:

The answer is: To accelerate an object <u>the force applied to the object</u> has to increase.

Explanation:

the acceleration of an object <u>increases with increased force</u> and <u>decreases with increased mass.</u>

3 0
3 years ago
gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are
AveGali [126]

Answer:

\lambda=1.03\times 10^{-3}\ m

Explanation:

Given that,

The energy of the microwave oven is 1.2\times 10^{-3}\ eV.

We need to find the wavelength of these photons.

1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J

The energy of a wave is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}

Put all the values,

\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m

So, the wavelength of these photon is 1.03\times 10^{-3}\ m.

7 0
3 years ago
In which city rusting is a problem delhi or mumbai
ss7ja [257]

Answer:

Mumbai

Explanation:

Because Mumbai is a coastal city

due to high humidity,vigorous rusting damages all the iron structures

∴ rusting is a major problem in Mumbai

3 0
3 years ago
Compare the properties of Titan’s atmosphere with those of Earth’s atmosphere.
Aloiza [94]

Answer and Explanation:

Comparison between the Titan's atmosphere and earth atmosphere

  • Titan atmosphere is more denser than the earth atmosphere
  • The quantity of nitrogen is more in titan atmosphere than earth atmosphere titan atmosphere have about 98 % of nitrogen in the other hand earth atmosphere has only 78 %
  • There are no oxygen present in titan atmosphere while in earth atmosphere it is present
  • Its air is not suitable for breath but in earth atmosphere we can breath
5 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
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