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3 years ago

About how much longer can the Sun continue to generate energy by nuclear reactions in its core?

Physics

1 answer:

QveST [7]3 years ago

5 0

Answer: 5billion years

Explanation: The sun produces energy through radioactive fusion reaction.

Nebula theory states that the gaseous particles of the Earth collapsed as a result of its own gravity which continuously lead to fusion reaction for the production of nuclear energy.

The Core of the Sun is that area up to 25% from the radius of the sun,here the pressure here range up to 250million atmosphere containing mainly hydrogen which gets converted in Helium molecule. The core is the center for energy production accounting for more than 98%, nuclear energy is transmitted at about 4.3million metric tons per second.

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Differentiate between rest and motion.

Lyrx [107]

Answer:

1. A body is said to be at rest if its position does not change with respect to its surroundings.

8 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

5,6,1,9,21,37,68 ? find missing

maria [59]

Answer:

Missing numbers are 3,11,13, 15,17,19, 23,27,29 and wrong are 6,68

4 0

3 years ago

40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker a

lakkis [162]

Answer:

Explanation:

there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.

7 0

2 years ago

A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor di

Marrrta [24]

Answer:

2625 m deep

Explanation:

Let the sound speed in sea water be 1500 m/s. If he hears the echo 3.5s after the strike, then the sound would have traveled a distance of 1500 * 3.5 = 5250 m to the bottom and back. This would mean the ocean is 5250 / 2 = 2625 m deep.

5 0

3 years ago