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Use the total area of the legs. The weight is distributed over the 4 legs.

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.

2 year old topic. I don't think it's relevant anymore.

That looks right. The negative image distance you got before implies that the film needed to be outside the camera to capture the image.

This is a 2 and a half year old topic.:wink:

Why did you reason that the image distance was negative?

You can have different compounds like ionic or organic and they have two different naming schemes.

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.

θ=ω_0 t+1/2 αt^2 then n=theta/2pi θ=0.5*pi*0.2+.5*1.8*pi*.04=pi(0.1+0.036)=0.136*pi θ/2*pi=0.136*pi/2*pi=0.136/2=0.068 rev

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.

Look at the derivative of e^kt and see how that relates to e^kt.

No, angular acceleration is 0.9 rad/s2. ac=vt2/r at=\alphar

Those are the only variables you need to know. I edited my above post, sorry for reposting what you knew.

\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

Yea those are the answers I got as well. Resultant vector depends on the tangential and centripetal accelerations.

I don't think you're calculating them correctly because they agree with the values I got which do not agree with your values.

If you set the surface of the water to h=0 then there will not be any remaining potential energy when the bucket hits the water.

B is right, the rest are wrong.

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?

Angular acceleration is constant. How did you end up with the 5.5 value?

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.

Yes, the moment of inertia formula for a disk and a cylinder are the same as long as the axis of rotation is through the center.

It states it is behind a 56mm focal length lens. Not 68mm behind the focal length of a 56mm focal length lens.

Why is di=68+56?

Yes, now how does that relate to work in and finally, work out.

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?

Well, 1/6*tan^-1(1)-1/6*tan^-1(0)=1/6*tan^-1(1)-0=1/6*tan^-1(1). So I guess you would need to know that tan^-1(1)=\pi/4.

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))

U substitution isn't needed. Look at an integration table. Hint: The function being integrated takes the shape of 1/(x2+a2)

Angular momentum is a vector. Flipping its direction has the effect of changing its sign in the equation.

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.

They could be giving you extra info for you to sift through and see what's relevant and what's not.

That's what I can't get around with this problem. Would the context come from what he did in class?

There you go guys. I said it was improper because of the notation, then I took it back. Sorry to offend you guys by saying something stupid.

First sum up all of the forces for the block. \SigmaFx=sin\theta*Fg-Ff-T=ma Then get the pulley \Sigma\tau=I*\alpha

Use the relationship \alpha=a/r and solve the torque equation for a.

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.

I said S was an improper subset.

Look at post #3, which has been said a few times now.

Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.

S is defined as an improper subset of V; so if V is a vector space, S must be as well.

Your reasoning is incorrect. Any identity matrix can be multiplied by a scalar to produce a singular, row equivalent 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?

Perhaps this would be useful. xm*xn=xm+n

I don't know how to do with algebra but, here's a hint: What two numbers can be raised to any power and still equal themselves?

\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.

yp=c3x2e-2x yp=(5/2)x*x2e-2x yp=(5/2)x3e-2x

(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