Recent content by Zero
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Calculating Electric Field: A Failed Attempt
Wow, thanks for the help everyone and sorry for my incompetence physics is not my strongest subject.- Zero
- Post #12
- Forum: Introductory Physics Homework Help
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Calculating Electric Field: A Failed Attempt
Ok i'll start guessing what it means, tell me if I'm correct, please. F is equal to 0 since the forces are balanced?- Zero
- Post #10
- Forum: Introductory Physics Homework Help
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Calculating Electric Field: A Failed Attempt
Ok I'm finally understanding what you are trying to say but I still have one more question, wheres the F value in the statement given?- Zero
- Post #8
- Forum: Introductory Physics Homework Help
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Calculating Electric Field: A Failed Attempt
Unfortunately, I don't understand what you are saying. At the moment my class hasn't gone over Lorentz Force (at least I don't think we did) so I don't think that would apply. Also for the problem where would, I start or what value can I get from what I have. Lastly, I looked up Lorentz force...- Zero
- Post #5
- Forum: Introductory Physics Homework Help
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Calculating Electric Field: A Failed Attempt
Yes, you are correct but that is not the main issue here. The thing I have problems with is the many unknowns present in the question. If you look at the relevant formula then it would be easy to see that based on the question as far as I can see we don't have enough variables to start. My...- Zero
- Post #3
- Forum: Introductory Physics Homework Help
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Calculating Electric Field: A Failed Attempt
I tried getting E by dividing volts and distance since I know the distance between the two plates is .352 m but it did not work- Zero
- Thread
- Electric Electric field Field
- Replies: 12
- Forum: Introductory Physics Homework Help
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Calculating the force due to friction
Thank you got what I was looking for really appreciate it.- Zero
- Post #3
- Forum: Introductory Physics Homework Help
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Calculating the force due to friction
p man= mv p man = 90 (7) p man= 630 p sled= mv p sled= 100 (0) p sled= 0 p man+sled= 630 + 0 p man+sled= 630 fg man= mg fg man= 90 (9.8) fg man= 882 fg sled= mg fg sled= 100 (9.8) fg sled= 980 fg sled+man= 980+882 fg sled+man= 1862- Zero
- Thread
- Force Friction
- Replies: 5
- Forum: Introductory Physics Homework Help
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A Spring pushing an object up a 25 cm slope
its all good man i should have also picked up on this too. Thank you I got the answer.- Zero
- Post #7
- Forum: Introductory Physics Homework Help
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A Spring pushing an object up a 25 cm slope
So what your saying is that its a trap, if so then I wasted hours on nothing. Thanx man really appreciate the help.- Zero
- Post #5
- Forum: Introductory Physics Homework Help
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A Spring pushing an object up a 25 cm slope
I don't know if I'm doing something wrong but I keep getting a negative number when I am left with v^2 which I can't square root. 1/2mv^2= .27 v^2 and mgh-1/2mv^2= -5.427 so therefore i get v^2=-20.1 which i can't square root. Can i just switch which one comes first then the answer would be...- Zero
- Post #3
- Forum: Introductory Physics Homework Help
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A Spring pushing an object up a 25 cm slope
f= kx = 150 n/m (.3m) = 45n Ug= 1/2 kx^2 = 1/2 (150) (.3)^2 = 75 (.09) = 6.75 j Ek= 1/2mv^2 6.75= 1/2 (.54) (v^2) 6.75= .27 (v^2) 25= v^2...- Zero
- Thread
- Slope Spring
- Replies: 6
- Forum: Introductory Physics Homework Help
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What is the equation for the object's velocity after being pushed in spring?
Got the answer thank you all for the effort- Zero
- Post #9
- Forum: Introductory Physics Homework Help
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What is the equation for the object's velocity after being pushed in spring?
Got the answer thank you all for the effort- Zero
- Post #8
- Forum: Introductory Physics Homework Help
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What is the equation for the object's velocity after being pushed in spring?
isn't initial velocity 0 so it wouldn't matter. Yes i know the kinetic energy is 30j and the velocity is 1.73 m/s but i just don't know how to get the answer for velocity unless you did a different formula than Ek=1/2mv^2 to find kinetic energy then that is possible to deduce that way- Zero
- Post #6
- Forum: Introductory Physics Homework Help