All categories

4 years ago

_____ describes the life cycle of a star. nuclear fusion nebular clouds stellar evolution the big bang

Physics

2 answers:

EleoNora [17]4 years ago

6 0

Stellar Evolution is the answer! :)

Anastasy [175]4 years ago

5 0

Answer:

Stellar evolution

Explanation:

Nuclear fusion is the process that goes inside a star which powers the star.

Nebular clouds are just clouds of dust (mostly hydrogen and helium) from which a star is born.

Stellar evolution describes the various stages in the life of a star.

The big bang describes how the universe as we know it actually is created. It says universe is created by the explosion of the tiny and highly dense singularity .

You might be interested in

What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?

serg [7]

Answer: $6.268(10)^{-16}J$

Explanation:

The kinetic energy of an electron $K_{e}$ is given by the following equation:

$K_{e}=\frac{(p_{e})^{2} }{2m_{e}}$ (1)

Where:

$K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}$

$p_{e}$ is the momentum of the electron

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

From (1) we can find $p_{e}$ :

$p_{e}=\sqrt{2K_{e}m_{e}}$ (2)

$p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}$

$p_{e}=2.091(10)^{-24}\frac{kgm}{s}$ (3)

Now, in order to find the wavelength of the electron $\lambda_{e}$ with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

$\lambda_{e}=\frac{h}{p_{e}}$ (4)

Where:

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

So, we will use the value of $p_{e}$ found in (3) for equation (4):

$\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}$

$\lambda_{e}=3.168(10)^{-10}m$ (5)

We are told the wavelength of the photon $\lambda_{p}$ is the same as the wavelength of the electron:

$\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m$ (6)

Therefore we will use this wavelength to find the energy of the photon $E_{p}$ using the following equation:

$E_{p}=\frac{hc}{lambda_{p}}$ (7)

Where $c=3(10)^{8}m/s$ is the spped of light in vacuum

$E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}$

Finally:

$E_{p}=6.268(10)^{-16}J$

4 0

4 years ago

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

3 years ago

How many joules of work are done on an object when a force of 10 N pushes it 5 m?

zhenek [66]

Answer:

option C

Explanation:

given,

Force on the object = 10 N

distance of push = 5 m

Work done = ?

we know,

work done is equal to Force into displacement.

W = F . s

W = 10 x 5

W = 50 J

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0

4 years ago

After the big bang, atoms in gas clouds experienced a greater gravitational pull to each other than atoms in other regions of th

allsm [11]

Answer:
These are the two statements with scientific facts that explain the described phenomenon
<span>
Gravitation between two objects increases when the distance between them decreases.</span>

When the mass of an object increases, its gravitational pull also increases.

Justification:

Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>

That law is represented by the equation:

F = G × m₁ × m₂ / d²

The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.

The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.

6 0

3 years ago

A spring of force constant 1500 Nm-l is acted

olchik [2.2K]

Answer:

1.876 J

Explanation:

First, let’s calculate the compression of the spring from the Hooke’s law:

F=kx,

here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.

Then, we get:

x=F/k=(75 N)/(1500 N/m)=0.05 m.

Finally, we can find the potential energy stored in the spring:

PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.

correct my answer if it's wrong ^^

7 0

3 years ago