3 years ago

Why would citizens need to be wary of debris on the road?

Physics

1 answer:

Feliz [49]3 years ago

8 0

In case they get affected by it

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i think its technician A

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Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the co

ElenaW [278]

Answer:

λ = 65.6 pm

Explanation:

Given that

λo = 65 pm

The initial energy of the electron

$E_o=\dfrac{hC}{\lambda_0 }$

Now by putting the values

$E_o=\dfrac{hC}{\lambda_0 }$

$E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}$

$E_o=3.05\times 10^{-15}\ J$

$E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV$

Eo=19.06 KeV

Given that kinetic energy KE= 0.84 KeV

Therefore the final energy

E= Eo - KE

E = 19.06 - 0.84 KeV

E= 18.22 KeV

The wavelength λ can be find as

$E=\dfrac{hC}{\lambda}$

$\lambda=\dfrac{hC}{E}$

$\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}$

λ = 6.56 x 10⁻¹¹ m

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3 years ago

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 83 vol% carbide

velikii [3]

Answer:

k_max = 31.82 w/mk

k_min = 17.70 w/mk

Explanation:

a) the maximum thermal conductivity is given as

$k_{max} = k_m*v_m + k_p*v_p$

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.17

v_p = proportion of carbide in the cement = 0.83

= 66*0.17 + 28*0.83

k_max = 31.82 w/mk

b) the minimum thermal conductivity is given as

$k_{min} = \frac{k_{carbide}* k_{metal}}{k_{metal} v_{carbide} +k_{metal} v_{carbide}}$

= \frac{28+66}{28*0.17 +66*0.83}

k_min = 17.70 w/mk

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