Answer:
a) 
For this case we know the following values:
So then if we replace we got:
b) 
With 
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
c) 
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
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:
Answer:
The mass will accelerate. Balanced Forces: When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. A plane will fly at constant velocity if the acceleration is zero.
Explanation:
Answer:
Explanation:
given.
magnification(m) = 400 x
focal length (f_0)= 0.6 cm
distance between eyepiece and lens (L)= 16 cm
Near point (N) = 25 cm
focal length of the eyepiece (f_e)= ?
using equation
780 seconds, or 13 minutes.
In the future, please use proper capitalization. There's a significant difference in the meaning between mV and MV. One of them indicated millivolts while the other indicates megavolts. For this problem, I'll make the following assumptions about the values presented. They are:
Total energy = 1.4x10^11 Joules (J)
Current per flash = 30 Columbs (C)
Potential difference = 30 Mega Volts (MV)
First, let's determine the power discharged by each bolt. That would be the current multiplied by the voltage, so
30 C * 30x10^6 V = 9x10^8 CV = 9x10^8 J
Now that we know how many joules are dissipated per flash, let's determine how flashes are needed.
1.4x10^11 / 9x10^8 = 1.56E+02 = 156
Since each flash takes 5 seconds, that means that it will take about 5 * 156 = 780 seconds which is about 780/60 = 13 minutes.
Answer: option D. the ratio of the population of male deer is not constant.
Explanation:
The bar graph permits to compare the results for two different populations: male and female deer in a very easy visual way.
These features are remarkable:
- The polulation of male deer (blue bars) decrease from 1961 to 1971, then increase in the next 10 year, decrease in the next decade, and increase for the next two decades. So, its trend is erratic, with ups and downs.
This discards the option A, which states that the population of male deer increases each decade from 1961 to 2011.
- The population of female deer (purple or brown bars) decreases every decade.
This discards the option B. which states that when the polulation of male deer increases, the poluplation of female deer also increases.
- The populations never are equal, hence this discards the option C.
- Since, one popultion increases and decreases, while the other population only decreases, you conclude that the ratio of the population of male deer to female deer is not constant, which is the option D.
