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Y_Kistochka [10]
3 years ago
10

The energy (such as light and heat) released by the Sun is produced by nuclear reactions in the core of the Sun, which convert h

ydrogen into helium. According to Einstein's equation E = mc2, the mass of one helium nucleus should be ___________.
Physics
1 answer:
Yanka [14]3 years ago
3 0

Answer:

<em>the mass of one helium nucleus should be</em> <em>less than the mass of four hydrogen nuclei.</em>

Explanation:

Deep inside the core of the Sun, enough protons can collide into each other with enough speed that they stick together to form a helium nucleus and generate a tremendous amount of energy at the same time. This process is called nuclear fusion.

The mass-to-energy conversion is described by Einstein's famous equation:

E = mc2, or, in words, energy equals mass times the square of the velocity of light. Because the velocity of light is a very large number, this equation says that lots of energy can be gained from using up a modest amount of mass.

Photons In the proton-proton chain reaction, hydrogen nuclei are converted to helium nuclei through a number of intermediates. The reactions produce high-energy photons (gamma rays) that move through the "radiative layer" surrounding the core. This layer takes up 60 percent of the radius of the Sun. It takes a million years for energy to get through this layer into the "convective layer", because the photons are constantly intercepted, absorbed and re-emitted. In the core, the helium nuclei make up 62% of the mass (the rest is still hydrogen). The radiative and convective layers have about 72% hydrogen, 26% helium, and 2% heavier elements (by mass). The energy produced by fusion is then transported to the solar surface and emitted as light or ejected as high-energy particles.

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The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no
Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?

From equation of normal strain in x direction:

\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta

Substituting the values:

\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}

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3 years ago
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garik1379 [7]
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6 0
3 years ago
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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

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Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

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\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

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Nastasia [14]

Answer:C

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