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kondaur [170]
3 years ago
5

Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres

, and that there are 5.0 cm3 in a teaspoon, what is the thickness (in cm) of a later of oil?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

 

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

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And replacing we have:

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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

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