The x-values would be (##-∞,∞##). Then it would be correct to say that there will not be a minimum and maximum, but there would be an infimum and supremum. ##inf S_3 = -∞## and ##sup S_3 = ∞##
##S_3## will then not be bounded above or below because the x-values are from negative infinity to...
I'm sorry about the ##x^2##, that was a typo. It should be ##(x + \frac 1 2)^2 ≥ -\frac 3 4##
If I look at the graph of ##y = x^2 +x +1##, I notice that the turning point of the parabola is at (-1/2 , 3/4).
Does this mean that ##S_3## is bounded from below? If it is bounded from below, does...
##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##
I am not sure if I have done this correctly. The infimum/supremum and maximum/minimum are confusing me a bit.
This is how I started:
##x^2+x+1=0##
##x^2+x+ \frac1 4\ =\frac{-3} {4}\ ##
## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ = 0##...
I tried to work out both a) and b), but I am not sure if I am correct. I drew a picture with a sphere around q first with radius r and then with radius 3r.
For a) ##E.A=\frac {q}{ε_°}## (when using Gauss' Law)
Since ##A=4πr^2##, I substituted this in the equation and solved for E giving me...
For ##a_r## I will use ##-b\dot \Theta^2##
##\vec T = -Mg\hat r## so that will be my 'force'.
$$-Mg = m(-b\dot \Theta^2)$$
Since ##u=b\dot \Theta## I can say that ##\dot \Theta = u/b##
Then: $$Mg = mb(u^2/b^2)$$
$$Mg = mu^2/b$$
##Mgb = mu^2## and therefore ##u^2 = Mgb/m##
I finally got it...
Yes, ##\ddot \Theta = 0## and if it is equal to zero, then ##\dot \Theta## is a constant. So Particle P has constant speed. Now I can say that ##u = b\dot \Theta##
Can I then say that the centripetal force is: $$F_C = mu^2/b$$
I do not see a nonzero ##\Theta## component on the three forces. So that means that ##\sum F_Θ = 0##
The Θ-component of ##\vec a## is the second part: ##b\ddot Θ \hat Θ##
Then ##\sum F_Θ = 0 = m\vec a_Θ = m(b\ddot Θ \hat Θ)##
I am a bit lost with the ##\hat \Theta## direction, but should it be: $$-Mg = m[(-b\dot Θ^2) \hat r + b \ddot Θ \hat Θ]$$
How do I show that ##\dotΘ## is constant? If it is constant then ##\ddot Θ=0## and the equation becomes ## -Mg = m(b\dot Θ^2 \hat r)##
OK. See if you can set up Newton's second law for the ##\hat \Theta## direction and try to use it to show that P has a constant speed.
After that, you can set up Newton's second law for the ##\hat r## direction.
For the ##\hat \Theta## direction, shouldn't it be the ##\vec v = b\dot Θ^2\hat...
I'm struggling to grasp the whole part about extent of dissociation. I know it should be the molecules that have dissociated, but this question is way different from what my book is saying. For the Change part in the ICE Table, my book states that the reactants should be -nα where n is the...
I was following the example in my book.
So then:
$$CO(g)$$
$$H_2O(g)$$
$$H_2(g)$$
$$CO_2(g)$$
Initial (mol)
0.40
1.00
0
0
Change (mol)
-x
-x
x
x
Equilibrium (mol)
0.40 - x
1.00 - x
x
x
a) The total equilibrium amount would then be n=(0.40 - x + 1.00 - x +x +x) mol = 1.40 mol
b)...
It should be the normal force. $$\vec F_N = mg\hat k$$ I think it should be positive since it is a force acting upward.
##\vec F_{grav} = -mg \hat k##.
Yes.
No. You already know the magnitude of ##T## from your consideration of particle Q. Can you use one of the unit vectors ##\hat k##, ##\hat...
So for the Change Part:
If 0.40x of CO reacts, then 0.40x of $$H_2O$$ also reacts?
Then, for the hydrogen gas and carbon dioxide, would it also be 0.40x (under the Change row)?
This is the free-diagram and the picture that I have drawn. I see did not include the i and j axis on the picture.
The two forces that are perpendicular to the table are the $$\vec F = -mg$$ and the unit vector $$\hat k$$
I think the third force would be the tension, T, on the string? Would the...