D. March because it is just below the 1 million marker on the graph and it is the only one that low.
        
             
        
        
        
Answer:
  y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
          1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
            1 / i = 1 / f - 1 / o
            1 / i_red = 1 / f_red - 1 / o
            1 / i_red = 1 / 19.57 - 1/30
            1 / i_red = 1,776 10-2
            i_red = 56.29 cm
Blue light
             1 / i_blue = 1 / f_blue - 1 / o
             1 / i_blue = 1 / 18.87 - 1/30
             1 / i_blue = 1,966 10-2
             i_blue = 50.863 cm
Now let's use the magnification ratio
              m = y ’/ h = - i / o
              y ’= - h i / o
Red Light
             y_red ’= - 5 56.29 / 30
             y_red ’= - 9.3816 cm
Light blue
             y_blue ’= 5 50,863 / 30
             y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
             y_red ’/ y_blue’ = 9.3816 / 8.47716
             y_red / y_blue = 1,107
             y_red / y_blue = 1.11
 
        
             
        
        
        
Answer:
1.3 x 10^(-2) atm/s
Explanation:
It follows the stoichiometry. For every mole of O3 that disappears, 1.5 moles (that is, 3/2) of O2 appears:
1.5 * 0.009 atm/s = 0.0135 atm/sec; the answer is 1.3 x 10^(-2) atm/s
 
        
             
        
        
        
A) an object with mass > 0 in a gravitational field
b) an object with an electric charge not 0 in an electric field
c) a moving object with an electric charge not 0 in a magnetic field