What is the weight on earth of a girl with a mass of 23kg?

If you see a sedimentary rock outcrop and red layers of sand are on top of pale yellow layers of sand, what do you know for sure

neonofarm [45]

Answer: The yellow layer is definitely older than the red layer

Explanation: According to Nicolaus Steno's law of superposition and original horizontality. Older rocks underlie younger rocks.

Sedimentary rocks are usually deposited in horizontal layers in which each stratigraphic layer is laid down before another can be deposited upon it.

The red layer, in addition to being older, is also likely to have undergone intense oxidation due to earlier exposure.

5 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

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Help! How do you find the Eccentricity??

denis23 [38]

Explanation:

The formula to determine the eccentricity of an ellipse is the distance between foci divided by the length of the major axis

8 0

2 years ago

Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.

Alika [10]

Let k = the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
= 0.5*(k N/m)*(0.35 m)²
= 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer: 53 N/m (nearest integer)

3 0

3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 119m in 29s?

Assoli18 [71]

Do 112m /29s which it will be 3.862 which if you round it, it will be 3.86 m/s

7 0

3 years ago

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