You have a distance and a time so you can work out the bikes speed which would be distance/time so 40mph.
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.
No, according to many pictures taken in space, the moon is white. However, on rare occasions, the moon appears blue.
Hope this helps! ☺♥
Answer:
27 m/s
Explanation:
Given:
v₀ = 15 m/s
a = 3 m/s²
t = 4 s
Find: v
v = at + v₀
v = (3 m/s²) (4 s) + (15 m/s)
v = 27 m/s
Answer:
A) The space time coordinate x of the collision in Earth's reference frame is
.
B) The space time coordinate t of the collision in Earth's reference frame is

Explanation:
We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).
An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.
<em>Lorentz transformation</em>
The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.
The Lorentz transformation is




prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations




First we calculate the expression in the denominator


then we calculate t




finally we get that

then we calculate x






finally we get that
