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Bad White [126]
3 years ago
6

This is the simplest type of fold where the rock bends like this:

Physics
2 answers:
Natasha_Volkova [10]3 years ago
6 0

The answer is Monocline. And I checked it, it's correct.

neonofarm [45]3 years ago
6 0

Answer:

The correct answer is option D, Monocline

Explanation:

Monocline is the simplest bend in the rock. It is also known as “one step” bend in the rock layer. Its structure is also very simple as the oldest rock lie at the bottom while the youngest rock lie at the top. It is a single , sharp bend which connect strata of different level during upward propagation of an extensional fault

.

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Answer: They have different rigidities.

Explanation:

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Water freezes at the freezing point to ice then melts to the melting turning it to liquid and vapor causing gas in precipitation
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A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t
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Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

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v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt

v_ave = v0+a(T/2)

We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

x(t) = x_A + v_0 t + \frac{1}{2}at^2

Where x_A is the position of point A. Therefore, the car will be at:

<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>

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.

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