Recent content by coolguy9
-
C
Falling Velocity Relative to Weight
The website is telling me 2.3 is incorrect.. I re-calculated, and it's coming up 2.3172, rounded to the tenths is 2.3.. Is there anything else we might not be accounting for?- coolguy9
- Post #9
- Forum: Introductory Physics Homework Help
-
C
Falling Velocity Relative to Weight
Ok, so.. Total PE = -1.96 -1.96 = (1/2) (0.49 + 0.24)(v^2) Assume the negative sign is negligible? v = 2.3?- coolguy9
- Post #7
- Forum: Introductory Physics Homework Help
-
C
Falling Velocity Relative to Weight
Alright, I'm still a little confused but I'm getting there, I think.. PE(mouse) = (9.8m/s)(.24kg)(0.8m) = 1.8816 PE(rat) = (9.8)(.49)(-0.8) = -3.8416 change in PE = 5.7232 5.7232 = 1/2 mv^2 v = 3.38? Which is still wrong, so.. I'm doing something wrong hehe- coolguy9
- Post #5
- Forum: Introductory Physics Homework Help
-
C
Falling Velocity Relative to Weight
I'm not making too much headway on this problem.. I've got PE = change in mgh and KE = change in 1/2mvf^2 - 1/2mvi^2 PE = 5.7232, assuming m = 0.490 kg + .240 kg, and g = 9.8, and h = .80m but I don't really know where to go from there. and is that PE even correct? or do I have to...- coolguy9
- Post #3
- Forum: Introductory Physics Homework Help
-
C
Falling Velocity Relative to Weight
Homework Statement A wooden rod of negligible mass and length 80.0 cm is pivoted about a horizontal axis through its center. A white rat with mass 0.490 kg clings to one end of the stick, and a mouse with mass 0.240 kg clings to the other end. The system is released from rest with the rod...- coolguy9
- Thread
- Falling Relative Velocity Weight
- Replies: 9
- Forum: Introductory Physics Homework Help