An object is moving along a straight line at a

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

λ = 65.6 pm

3 0

3 years ago

A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter

lutik1710 [3]

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72 m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion $(v_t=\omega\,R=\omega\,(0.25\,m)\,)$ - with $\omega$ being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals: $\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}$