Recent content by rocky4920

I never said I was angry or mad at you or the Praxis... I simply made this post to ask for some guidance towards the Praxis. I did not know that this would turn into the vetting of my past posts on here. I was under the impression the goal of this forum is supposed to be about helping out...

I agree, and I have been focusing on my areas of weakness from my practice test results.

I graduated from college back in 2013 and had 15 credits. Yes, in 2017-2019 I took additional online classes to get me to 30 credits. "I don't understand how you were able to solve advanced problems a year or two ago and can't solve simpler problems now." Well, I didn't publish my post to be...

Hello, I have been having trouble successfully passing the Physics Praxis 5265 Exam. I am a high school biology teacher and have 30 college credits in Physics. My district wants me to teach Physics and I need to get my certification. Once I pass the praxis I will be able to teach. I have...

Photo:

FE and FB. FE-FB= zero. The electric field and the magnetic field.

Thank you for the help. Yes, I definitely was using the wrong equation. F= ILA, therefore, I= F/LB Is that a better approach to use?

Based on the right hand rule, though, wouldn't the field go out of the page above the wire, and then into the page below it?

So by doing so I would manipulate the following formula: B = µo I/2 pi r ? I = 2 pi r B / [FONT=PT Sans] µo r = 0.25m since wire is .50 m B= 1.6 T µo= 4 pi x 10-7

Homework Statement Homework Equations The Attempt at a Solution [/B] My question: in order to find the answer for part d, I have the L, the B, and the mass. However, the F equations have I, which I do not have. Would I have to solve for I in order to find F? If so, I would like some...

V= IR V= 1.5 (3) V= 4.5 V for voltage drop for 3 ohm resistor.

Ok so would the current for R1 R2 be the same for R3 R4 being 1.5 amps? Therefore, making R1, R2, R3, and R4 V= IR V= (5 ohms) (1.5 A) V= 7.5 V at each throughout circuit.

There was no battery voltage given in the problem.

Sorry about that! Image above.