Recent content by MissPenguins
-
M
Calculating the Net Charge Enclosed by a Closed Surface
Alright, thank you very much. My professor will probably explain it tomorrow. Thanks everyone.- MissPenguins
- Post #8
- Forum: Introductory Physics Homework Help
-
M
Calculating the Net Charge Enclosed by a Closed Surface
Is it calculating the area of the cube? Can you please explain the concept? I am watching a youtube video on electric field now. Thanks.- MissPenguins
- Post #5
- Forum: Introductory Physics Homework Help
-
M
Calculating the Net Charge Enclosed by a Closed Surface
So electric flux = E dot dA?? Can you please explain the concept? Thank you!- MissPenguins
- Post #3
- Forum: Introductory Physics Homework Help
-
M
Charges mass on the end of a light string
I figured the problem out. Thank you very much everyone!- MissPenguins
- Post #9
- Forum: Introductory Physics Homework Help
-
M
Charges mass on the end of a light string
You said E is different for part 2 since the charge density changed. So how do I find E? I attempted to do (0.25E-6)/(2*8.854E-12) = 14117.91281 for E, is this right for E? Then use (14117.91281)(2*8.854E-12) = 2.5E-7? It doesn't make sense. Can you give me more hints? Thanks.- MissPenguins
- Post #7
- Forum: Introductory Physics Homework Help
-
M
Charges mass on the end of a light string
The previous one wasn't a mistake. It's qE = mgtan(77) = 0.042448 I already did this \sigma = (E)(2Eo). So (0.042448)(2(8.85X10-12) = 7.5133X10-13. I converted back to uC = 7.51633X10-7. And it is wrong! :(- MissPenguins
- Post #5
- Forum: Introductory Physics Homework Help
-
M
Calculating the Net Charge Enclosed by a Closed Surface
Homework Statement A closed surface with dimensions a = b = 0.294 m and c = 0.3528 m is located as in the figure. The electric field throughout the region is nonuniform and given by \vec{}E = (\alpha+\beta x2)ˆı where x is in meters, \alpha = 2 N/C, and \beta = 4 N/(Cm2). See figure...- MissPenguins
- Thread
- Charge Closed Net Surface
- Replies: 7
- Forum: Introductory Physics Homework Help
-
M
Charges mass on the end of a light string
Yea. tan(77) = (qE/mg) which is qE=mgtan(77) = 0.04244 I am not sure what other charge density do I have to use. E0is a constant, that's given. PLEASE HELP! Same question as this one https://www.physicsforums.com/showthread.php?t=109643 Except part 2 was not explained.- MissPenguins
- Post #3
- Forum: Introductory Physics Homework Help
-
M
Charges mass on the end of a light string
Homework Statement A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet (with areal charge density 0.25 μC/m2) of infinite extent. Look at attachment for diagram Find the angle \theta the thread makes with the vertically charged...- MissPenguins
- Thread
- Charges Light Mass String
- Replies: 8
- Forum: Introductory Physics Homework Help
-
M
Power of Wave: Calculating P & P0 Ratios
I submitted the answer, and it's wrong! :(- MissPenguins
- Post #5
- Forum: Introductory Physics Homework Help
-
M
Power of Wave: Calculating P & P0 Ratios
So is my approach correct?- MissPenguins
- Post #3
- Forum: Introductory Physics Homework Help
-
M
Power of Wave: Calculating P & P0 Ratios
Homework Statement A wave pulse traveling along a string of linear mass density 0.0043 kg/m is described by the relationship y = A0 e −b x sin(k x − ω t) , where A0 = 0.0032 m, b = 0.68 m −1 , k = 0.57 m −1 and ω = 44 s −1 . What is the power carried by this wave at the point x = 2.2...- MissPenguins
- Thread
- Power Wave
- Replies: 6
- Forum: Introductory Physics Homework Help
-
M
Calculating Oscillatory Motion Parameters for a Spring-Block System
Can you please tell me where did you get 4.375 s from? thanks. Nvm, I got it w = sqrt of k/m. Hehe, I got the right answer too, thank you very much.- MissPenguins
- Post #12
- Forum: Introductory Physics Homework Help
-
M
How Do You Calculate the Spring Constant for a Car's Shock Absorbers?
Alright, I figured it out and got the right answer. Thanks. ;)- MissPenguins
- Post #4
- Forum: Introductory Physics Homework Help
-
M
How Fast Can a Piston Move in Simple Harmonic Motion?
This part is wrong. Remember 1 rev = 2pi. So you multiply by 2pi, and then 1 min = 60 seconds. So you divide by 60 seconds. I did not get 73 rev/s. I got another number.- MissPenguins
- Post #6
- Forum: Introductory Physics Homework Help