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grigory [225]
2 years ago
12

Granite is mostly used in building work places whereas Gneiss is used for making tombstones, flooring, etc. Why do you think so?

Explain how does this happen in nature
Physics
1 answer:
lozanna [386]2 years ago
6 0

Answer:

Granite is durable, as it is hard and tough.

Gneiss has resistance to pressure and mechanical impacts

Explanation:

Granite is an igneous rock. It is mostly used in building works and construction because they are very durable. They are hard and tough and they have no internal structures.

Gneiss is used for flooring, ornamental stone, tombstones because of the fact that it shows resistances to pressure and also mechanical impacts.

<u>how they are formed in nature:</u>

In nature, granite is formed from the cooling down of hot molten magma and it's solidification before it reaches the surface of the earth.

In nature, gneiss is as a result of igneous rock or sedimentary rocks metamorphosing. Gneiss and granite are kind of similar. When subjected to great heat, granite becomes gneiss

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A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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=157^{\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-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

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

here, \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}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

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

8 0
3 years ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0
3 years ago
This illustration represents the compoundA)carbon oxide.B)carbon dioxide.C)carbon monoxide.EliminateD)monocarbon oxide.
pentagon [3]
B.) Carbon Dioxide because the carbon is surrounded by oxygen
3 0
3 years ago
Read 2 more answers
Physics!!! please help D:
Ugo [173]
Answer: A

Explanation: isotopes of the same thing element have the same number of protons in the nucleus but differ in the number of neutrons.
6 0
2 years ago
An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

8 0
2 years ago
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