Now you have the opposite problem. There's a negative sign in this one.
m= -v/u
The image gets flipped upside down. This happens with your eyes too; your brain flips the image back right side up.
Yup.
http://www.fairbornchempage.com/Resources/Prefixes.htm
Yes.
The term you were probably looking for was "chemical nomenclature" for the Google search. But you also need to know the different types of molecules to know how to name them.
a) correct
b) still incorrect
c) correct, m/s (rads are kind of like a place holder unit in this case so it goes away, i don't know the actual reason why though)
d) correct, m/s2 (for same reason)
a) You didn't convert the units of ω0.
b) Wrong for same reason as a.
c) Wrong for same reason as a.
d) Can you show the calculation?
You know what your doing, just didn't convert.
\Deltax=vit+\frac{1}{2}at2
There's your kinematic equation.
Edit: sorry didn't see your attempt.
\Deltax=vit+\frac{1}{2}at2
0=vit+\frac{1}{2}at2-\Deltax
\frac{1}{2}at2+vit-\Deltax=0
Quadratic form
Towards the left of the y-axis is -x direction and right of the y-axis is +x direction. So the x-component of one of the vectors is negative. Which one is it?
If they ask you for the torque you need to know the radius. In this case you just used the definition of torque without having to calculate torque itself. Problems sometimes give extra information to force you to think about what it useful info and what is not useful.
Work is Joules/s. 580 kg/s fall off the dam. So work in is the energy associated with the mass that falls each second. What's the energy associated with the mass?
This problem looks familiar did you post it on some other message board?
As you stated the mechanical energy is the ratio of the work out to the work in. What's the work in? The efficiency is 55% so, 55% of the work in is returned.
Nothing needs to be done with \sqrt{2}, it is a constant. What's the derivative of a constant? (btw, \sqrt{2} is an irrational number which is why you couldn't rationalize it.)
The coordinates are the x and y-values of the function and y=f(x) so, (x,y)=(x,f(x))
Looks good. The only thing that I would change is "Assume a < b and c < d". Assuming something does not mean it is necessarily correct as a proof by contradiction can demonstrate. So you can change it to something like "Stated a < b and c < d." Since a < b and c < d was given to you.
So it's false because row operations can be carried out on a derived row equivalent row echelon matrix to produce another row equivalent row echelon matrix.
When you say it is unique do you mean there is a one-to-one relationship with every matrix to it's reduced echelon form or that there is only one reduced echelon form for any matrix?
\mathbb{R}^n is an n-dimensional vector space. {x1,x2,...,xn} is a spanning set of \mathbb{R}^n of length n. This makes {x1,x2,...,xn} a basis of \mathbb{R}^n, which means it must be linearly independent.
(D2+4D+4)(c3x2e-2x) = 5xe-2x
c3(4x2e-2x-8xe-2x+2e-2x+4(2xe-2x-2x2e-2x)+4x2e-2x) = 5xe-2x
Every thing but c3(2e-2x) reduces to zero
c3(2e-2x)=5xe-2x
c3=(5/2)x