Uranium-235 has to be enriched before it can be turned into fuel rods. This enrichment is achieved by 1.___ and2. __

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

Nataliya [291]

Answer:

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 ( sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees

8 0

4 years ago

What is the correct order of the layers' density from lowest density to highest?

Orlov [11]

Answer:

C. crust, mantle, core

Explanation:

density increases as you travel from the crust to the inner core

the crust is on top

next is the mantle

and then the core

6 0

3 years ago

Read 2 more answers

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

4 years ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

ratelena [41]

Answer:

a = 1.3 m/s2; fn = 63.1 n :)

8 0

3 years ago

Calculate the rate of heat generation in kW due to the burning of the fuel when you drive a car rated at 34 MPG (miles per gallo

alexira [117]

Answer:

Explanation:

Let us calculate gallon used in one hour .

It travels 70 miles in one hour

in 70 miles it uses 70 / 34 gallons of fuel

70 / 34 gallons = 70 / 34 x 3.7854 kg

= 7.8 kg

heat generated = 7.8 x 44 x 10⁶ J

= 343.2 x 10⁶ J

This is heat generated in one hour

heat generated in one second = 343.2 x 10⁶ / 60 x 60 J/s

= 95.33 x 10³ J /s

= 95.33 kW.

4 0

3 years ago

Uranium-235 has to be enriched before it can be turned into fuel rods. This enrichment is achieved by 1.___ and2. ___

Uranium-235 has to be enriched before it can be turned into fuel rods. This enrichment is achieved by 1._ and2. _