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arlik [135]
3 years ago
15

PLEASE HELP ASAP!! FILL IN THE BLANKS. 99 POINTS!The force of _________ is what holds you down to earth. The ______________ of o

n a vector is indicated by the length of its arrow. A measure of the force of gravity on an object is called __________. Einstein explained gravity as a __________ in space. If you had the mass of 600 kg on the earth; you would have the mass of ______kg on the tiny Mercury.
Physics
2 answers:
WARRIOR [948]3 years ago
7 0

Answer:

Explanation:

The force of __gravity__ is what holds you down to earth. The __magnitude__ of on a vector is indicated by the length of its arrow. A measure of the force of gravity on an object is called __weight__. Einstein explained gravity as a __distortion__ in space. If you had the mass of 600 kg on the earth; you would have the mass of _600_kg on the tiny Mercury.

Dima020 [189]3 years ago
4 0

Answer:

The force of <u>Gravity</u> is what holds you down to earth.  

The <u>length</u> of on a vector is indicated by the length of its arrow.

A measure of the force of gravity on an object is called <u>weight</u>.

Einstein explained gravity as a <u>distortion </u>in space.

If you had the mass of 600 kg on the earth; you would have the mass of <u>226.8 kg</u> on the tiny Mercury.

I hope this helps you! :)

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