During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel?

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 many cartoon shows, a character runs off a cliff, realizes his predicament, and lets out a scream. He continues to s

IrinaVladis [17]

Answer:

Here the source is moving away from the observer so frequency will be smaller than the actual frequency and since the speed is increasing so the frequency is decreasing with time so correct answer is

D) lower than the original pitch and decreasing as he falls.

Explanation:

As we know by the Doppler's effect of sound we have

so we will have

$f = f_o(\frac{v}{v + v_s})$

so here when source moves away from the observer with a some speed then the frequency of the sound observed by the observer is smaller than the actual frequency

Here we know that the speed of the source is increasing with time as the source is falling under gravity

So we can say that the pitch of the sound will decrease with time

4 0

3 years ago

Please answer this im not sure

vovikov84 [41]

Answer:

the answer is B

8 0

3 years ago

When a body is moving with a uniform velocity, the acceleration is ___?

denis23 [38]

According to the formula :
a=ΔV / ΔT
When a body is moving with a uniform velocity, the acceleration is zero. That's it. You should remember, that velocity is not constant whereas speed is constant.

4 0

3 years ago

Which graph accurately shows the relationship between kinetic energy and speed as speed increases?

mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$