Can someone help me and put them in order, I numbered them down so it can be easier to say.

X rays of wavelength 0.0100 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

Fofino [41]

Answer:

a) $=4.84*10^{-12}$

b) $= -2.76*10^{-14} J$

c) $i.e -2.76*10^{-14} J$

d)= 0 and the direction of motion is equal to zero

Explanation:

a) compton shift

$\Delta\lambda = \frac{h}{mc} (1-cos\theta)$

$\Delta\lambda = \frac{6.626*10^{-34}}{9.11*10^{-11}3*10^8} (1-cos180)$

$=4.84*10^{-12}$

b) the new wavelength

$\lambda' = 10.0*10^{-12} +4.84^10^{-12}$

$=14.84*10^{-12} m$

$\Delta E = E' - E$

$=hc[\frac{1}{\lambda'}-\frac{1}{\lambda}]$

$\Delta E = 6.626*10^{-34}*(3*10^8)[\frac{1}{14.84*10^{-12}}-\frac{1}{4.8*10^{-12}}]$

$= -2.76*10^{-14} J$

C)By conservation of energy, the kinetic energy of recoiling electron is equal to the magnitude of energy between the photon energy

$i.e -2.76*10^{-14} J$

d) the angle between the positive direction of motion

$sin\phy = \frac{\lambda_t sin\theta}{\lambda'}$

$=\frac{2.43*10^{-12}sin180}{14.84*10^{-12}}$

= 0

the direction of motion is equal to zero.

4 0

4 years ago

The sun is 1.50x10^11 m from earth. How long does it take the suns light to reach earth? How long

galina1969 [7]

Answer:

what i don't understand the question

3 0

3 years ago

How high off the ground is a 5.5 kg object that has a gravitational potential energy of 8500 DE=mah

umka21 [38]

Answer:

h = 157.70 meters

Explanation:

Given the following data;

Mass = 5.5 kg

Gravitational potential energy = 8500 Joules

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the height of the object;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

8500 = 5.5*9.8*h

8500 = 53.9h

h = 8500/53.9

h = 157.70 m

7 0

3 years ago

If you needed information about a scientificconcept, whom would you ask? Check all thatapply.my teachermy parentsa university pr

allochka39001 [22]

A librarian... it might also be a scientist

4 0

3 years ago

Read 2 more answers

A flywheel with a radius of 0.200m starts from rest and accelerates with a constant angular acceleration of 0.900rad/s2.

kobusy [5.1K]

Answer:

The magnitude of the radial acceleration is 0.754 rad/s²

Explanation:

Given;

radius of the flywheel, r = 0.2 m

initial angular velocity of the flywheel, $\omega_i = 0$

angular acceleration of the flywheel, a = 0.900 rad/s².

angular distance, θ = 120⁰

the angular distance in radian = $\frac{120}{180} \pi = \frac{2 \pi}{3} \ rad$

Apply the following kinematic equation to determine the final angular velocity;

$\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2(0.9)(\frac{2\pi}{3} )\\\\\omega _f^2 = 3.7699\\\\\omega _f = \sqrt{ 3.7699} \\\\ \omega _f = 1.942 \ rad/s$

The magnitude of the radial acceleration is calculated as;

$\alpha _r = \omega ^2r\\\\\alpha _r = (1.942)^2 (0.2)\\\\\alpha _r =0.754 \ rad/s^2$

Therefore, the magnitude of the radial acceleration is 0.754 rad/s²

4 0

3 years ago