All categories

3 years ago

The law of conservation of momentum states that...

Physics

1 answer:

Brums [2.3K]3 years ago

4 0

Answer:

The momentum before is equal to the momentum after

Explanation:

It is equal and should level out in an equation.

You might be interested in

How much heat is needed to vaporize 25.0 kg of water (Ly - 540 kcal/kg)?

HACTEHA [7]

Answer:

4.- oh shoot my dad is calling me brb ok im back so i think the answer HOLY **** my girlfriend keeps nagging me

Explanation:

4 0

3 years ago

1. 3 main components of a circuit

siniylev [52]

Voltage, resistance and current are the three components that must be present for a circuit to exist. A circuit will not be able to function without these three components. Voltage is the main electrical source that is present in a circuit. :)

7 0

3 years ago

Read 2 more answers

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

In the decomposition of water, why is twice as much hydrogen as oxygen formed?

astra-53 [7]

Answer:

I think

D : Hydrogen gas has two atom , and oxygen has one atom

7 0

3 years ago

If opposite poles repel each other, why does the north end of a compass point to the north pole?

lina2011 [118]

Ratios are fractions

6 0

3 years ago