Recent content by chimpfunkz

I had a brain teaser that I was hoping you could solve: "In order to guarantee all emergency call are received, the phone company wants all phone nubers with 9 follower by 2 1's to be routed to 9-1-1 emergency center. This will ensure that any real emergency victims will still get help they...

Homework Statement You are running at 4.5 m/s around a circle 12m in diameter. Is the centripetal acceleration perceptible? Homework Equations a=v^2/r The Attempt at a Solution I know that the centripetal acceleration is 1.7m/s^2 (4.5^2/12). What I don't know is what it means for...

Wow, that makes it so much clearer. The only step I don't get is how you get from \frac{-\sqrt{2}vl}{v^2-2gl} to \frac{\frac{l\sqrt{2}}{v}}{\frac{2gl}{v^2}-1}

You were right, I did lose the square. It is actually t_b - t_c = 2 \frac{\frac{-gl^2}{v^2}}{\frac{v^2-2gl}{\sqrt{2}v}} - \frac{\sqrt{2}l}{v} And I know this might be a lot to ask, but could you take the formula above, and maybe explain quickly how to simplify it? Algebra is not really my...

Alright, so at long last, this is what I have: t_b-t_c=2\frac{\frac{-gℓ}{v^2}}{ \frac{v^2-2gℓ}{√2v}} - \frac {√2ℓ}{v} The question is, what does this simplify too? What should I do to make the algebra a little easier? Would the final result of this match the answer my teacher gave for...

It seems you have the idea. To find the final temperature, try setting -qcopper = (specific heat)(mass)(change in Temp.) equal to qwater = (specific heat)(mass)(change in Temp.) Another thing to keep in mind that should help you solve for final temperature, is that the at the end, the...

So for Tc, t_c=\frac{√2 ℓ-v_0t_b)} {v_0} So you plug that in for when y(t) is equal for both the baby and the cat, which is \frac{v_0}{√2}sin(0)t_c-\frac{1}{2}gt_c= v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt^2_b or, since sin(0)=0, -\frac{1}{2}gt_c= v_0sin(\frac...

whoops! I forgot my squares! so the correct equations y_b = v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt^2_b and y_c = \frac{v_0}{√2}sin(0)t_b-\frac{1}{2}gt^2_c But how could you simplify tc? would the simplest form just be t_c=\frac{√2( ℓ-v_0cos(\frac {π}{4})t_b)} {v_0} Then after this is where I...

Let me see if I am on the right track: y_b = v_0sin(\frac {π}{4})t_b-\frac{1}{2}gt_b and y_c = \frac{v_0}{√2}sin(0)t_b-\frac{1}{2}gt_c as well as ℓ = x_b + x_c or ℓ = v_0 cos(\frac {π} {4}) t_b + \frac{v_0}{√2}cos(0)t_c So when I solve for t_c using the formula for...

Homework Statement A cat and a baby are launched from skeet traps on opposite sides of the Grand Canyon, which are at the same height and a distance ℓ apart. The cat is launched horizontally at speed (v0/√2), the baby at initial speed v0 at an angle of (pi/4) above the horizontal. By amazing...