Recent content by BlueEight

Ah, thanks. But why would it be "necessary for experimental precision?"

Oh yeah! Because if it was horizontal, there would be no "vertical tension" holding up the mass of the stopper?

Homework Statement http://img.skitch.com/20091201-pxecs3d574itggpejybhdbp3qr.jpg Essentially, as an experiment, our physics class whirled a stopper attached to a string that went through a tube that was finally attached to a weight on the bottom of the string. The Attempt at a Solution...

From my limited knowledge, normal force is what slows down an object after impact. Therefore, according to Newton's Second Law, Normal force N = mg + ma. However, I do not know how the acceleration will be affected after impact, and thus declare this problem unsolvable, having only gone through...

You're referring to this, right? a=\frac{V^2-{V_{imp}}^2}{2\Delta{y}} Also, no. That only works for after he hits the ground. Read through my explanation again.

In these problems, if his initial velocity isn't give, assume he just walks off.

We DO know Vo. It's his INITIAL velocity when he starts falling. Imagine that he's a particle-like object that you're holding. What is his speed right when you drop the object?

What? To get Vimp, we use {V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}} Vo means initial velocity, aka his velocity when he first walks off/jumps off the ledge.

Yes. (If by V you mean Vo):smile:

You know all the "velocities" that you'll need. Granted, you need to infer it from the problem, but it's there.

You want a Lewis Dot Structure of AgNO3, not just NO3-.

Try reading http://www.purplemath.com/modules/polydiv2.htm" It has a good explanation and an example.

I realize that the gravitational potential energy must be greater, but why is is exactly twice, no matter the mass of the object?

When the bomb is dropped, it will have a velocity of 284 m/s, which will be SOLELY horizontal. Because of independent superposition, the fall of the bomb will not be affected by its horizontal movement. Therefore, solve for the time that the bomb will take to reach the ground as if it were in...

Hint: Use your kinematic equations for uniform acceleration.