Recent content by cherryrocket

Hey I finally get it! Thanks alridno and mace2!

http://i158.photobucket.com/albums/t88/liliananas/FORCES.jpg My teacher told us that on an inclined plane, the angle the weight of a block makes with the parallel force is the same as the angle of the inclined plane, in this case 29°. Why is that? I know it has to do something with similar...

I just finished a question in which a u-shaped tube (manometer) is filled with water to a certain height. Another substance, in this instance, oil, is added to one side of the manometer. I was asked to find the height of the new substance. I used the formula P(water) = P(oil), or (ϱwater)gh =...

Thanks everyone for your help!

Someone please answer my question. I really need it... it is the last question on my assignment and I need to understand it... thnks

Nancy's dropped camera REvised okay, so first I went: d=vt =4m/s(2s) = 8m. So 8m is the distance that the helicopter rised in 2 secs, from the time the camera was dropped. Then I look for the final velocity of the camera, because I will need it in a formula. Vf = 4m/s + (-9.81m/s^2 X...

Yes, I think that perhaps I should read over the whole chapter on Accelerated Motion before trying to do this module. Well, thanks anyway.

Confused! so, what you did was rearrange vi=d-1/2at^2 / t, right? d= (0m/s)(3.85s) + (1/2)a(3.85s)^2 = 0 + 7.41a = 7.41a but I still can't find a because I don't have the average velocity. I'm so confused...

7.Q: "Nancy was riding in a helicopter which was rising at 4.0 m/s when she dropped her camera out. After 2.0 s how far apart were the camera and the helicopter?" A: This isn't an acceleration question is it? Just velocity? d=vt = (4.0m/s)(2.0s) = 8.0m 8.Q: "A certain car tire can...

1. Q: "John slides down on a hill on his toboggan. He starts from rest and speeds up to 18.0 km/h in 8.0 s. Find his acceleration in m/s^2." A: Convert 18.0 km/h to m/s^2 = 18.0 km/h(0.2778) = 5.00 m/s a=v/t = 5.00m/s / 8.0s = 0.63 m/s^2 2.Q: Explain how you can accelerate without speeding...

I don't know, I just thought it was too easy to be the right answer... does that make sense? lol ps. how do you close a thread when a question is solved?

So then I did get the right answer, just didn't round properly, right? Thanks!

REVISION to simple velocity Q. ***REVISION*** Juanita left on a walk 45 min ago. She is traveling at 4.5 km/h. How fast must you travel to catch up to her in 1.0h"? Juanita t=45 min +1.0h min v=4.5 km/h d=vt = (4.5km/h)(1.75h) = 7.8 km Me - v=d/t = (7.8km/1.0h) = 7.8 km/h I must travel at...

[SOLVED] Another simple velocity question. "Juanita left on a walk 45 min ago. She is traveling at 4.5 km/h. How fast must you travel to catch up to her in 1.0h"? Juanita t=45 min v=4.5 km/h d=vt = (4.5km/h)(0.75h) = 3.375 km Me - v=d/t = (3.375km)/1.0h = 3.4 km/h I must travel at...

7. Q: "If you must average 60 km/h for a 90 km trip, and you only average 40 km/h for the first 40 km, what velocity must you average the rest of the way?" A: 90 km 60km/h 1.5h d v t 1st 40 km 40 km/s 1h d v t 2nd 50 km ...