<span>P = energy/t = 0.0025/1E-8 = 250000 W
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2
I(peak) = 2I(ave) = 8.8113463E11 W/m^2
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>
Answer:
The concentration of hydrogen ion at pH is equal to 2 :![= [H^+]=0.01 mol/L](https://tex.z-dn.net/?f=%3D%20%5BH%5E%2B%5D%3D0.01%20mol%2FL)
The concentration of hydrogen ion at pH is equal to 6 : ![[H^+]'=0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D0.000001%20mol%2FL)
There are 0.009999 more moles of
ions in a solution at a pH = 2 than in a solution at a pH = 6.
Explanation:
The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
The hydrogen ion concentration at pH is equal to 2 = [H^+]
![2=-\log [H^+]\\](https://tex.z-dn.net/?f=2%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C)
![[H^+]=10^{-2}M= 0.01 M=0.01 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2%7DM%3D%200.01%20M%3D0.01%20mol%2FL)
The hydrogen ion concentration at pH is equal to 6 = [H^+]
![6=-\log [H^+]\\\\](https://tex.z-dn.net/?f=6%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%5C%5C)
![[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6%7DM%3D%200.000001%20M%3D%200.000001%20mol%2FL)
Concentration of hydrogen ion at pH is equal to 2 =![[H^+]=0.01 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01%20mol%2FL)
Concentration of hydrogen ion at pH is equal to 6 = ![[H^+]'=0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D0.000001%20mol%2FL)
The difference between hydrogen ion concentration at pH 2 and pH 6 :
![= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L](https://tex.z-dn.net/?f=%3D%20%5BH%5E%2B%5D-%5BH%5E%2B%5D%27%20%3D%200.01%20mol%2FL-%200.000001%20mol%2FL%20%3D%200.009999%20mol%2FL)
Moles of hydrogen ion in 0.009999 mol/L solution :

There are 0.009999 more moles of
ions in a solution at a pH = 2 than in a solution at a pH = 6.
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Answer:
Explanation:
Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
Use third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Thus, the second rock reaches the 4 times the distance traveled by the first rock.
I think the answer is B true