X -> Y + 2Z
So there are 2 different particles. 1 mol of X produces
1 mol of Y and 2 moles of Z.
Kps = [Y] [Z]^2
We will call “s” (solubility) the molarity of X
So the molarity of Y+ is also “s” (same number)
And the molarity of Z is “2s” (twice as much)
Kps = s*(2s)^2 = s*4s^2=4s^3
If s is multiplied by 2:
Kps = 4*(2s)^3=4*2^3*s^3=4*8*s^3
So Kps is multiplied by 8.
Momentum = Mass x Velocity
Put the values where they belong and solve for Velocity.
In this case, since Mass is being multiplied by Velocity, to solve for be Velocity you would divide both sides by Velocity. The velocity will equal the momentum divided by the mass.
Answer:
34.70 N
Explanation:
Given :
F~ 1 = 43 N in direction 38◦
F~ 2 = 26 N in direction −140◦
F~ 3 = 27 N in direction 110◦
Therefore,
F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)
= 43 (0.7) + 26 (-0.7) + 27 (-0.3)
= 3.8
F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)
= 43 (0.6) + 26 (-0.6) + 27 (0.9)
= 34.5
so, F~ = 
= 34.70 N
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
