Recent content by saw176

Ah! Got it I think! I = 2 / (2.27 k-ohm) I = 8.81 x 10-4 Total resistance is still 11.67 k-ohm, so: Emf= 8.81 x 10-4 (11.67 k-ohm) Emf= 10.3 = 10 V Thanks for your help!

Thanks, that makes sense. So the actual resistance of the meter would be 3000 ohm then, if I use R/3V = 1000 Ohms/Volt. I took this resistance and repeated my process: I = 2 V/ 3000 ohm = 6.67 x 10-4A Total Circuit resistance = 1/9.4 kohm + 1/3000 ohm = 2.27 k-ohm 2.27 kohm + 9.4 kohm = 11.67...

Homework Statement Two 9.4 k-ohm resistors are placed in series and connected to a battery. A voltmeter of sensitivity 1000 ohm/V is on the 3-V scale and reads 2 V when placed across either resistor. What is the emf of the battery? (Ignore it's internal resistance) Homework Equations...

Oh okay I see, that makes sense, thanks gneill! Edit: I see Ouabache's reply now too, you guys are right in pointing out my mistake of assuming the the voltage drop was the voltage of the battery, I knew it seemed kind of fishy but didn't really know what I was doing. Thanks guys!

I think I got it! Okay, I went V = 5.25 x 10-3A (480 +750)ohm to get a battery voltage of 6.45 volts. I then added the 0.33 volts that the ammeter was reading to this, to get 6.78 volts. Then, I applied V=IR, 6.78 V = I (1230 ohm) to get the answer 5.52 x 10-3A. This is the correct answer, so I...

Homework Statement An ammeter whose internal resistance is 63 ohm reads 5.25 mA when connected in a circuit comtaining a battery and two resistors in series whose values are 750 ohm and 480 ohm. What is the actual current when the ammeter is absent?Homework Equations V = IR Rseries = R1 + R2The...

Thanks ehild! I actually saw somewhere online else a student saying they got that same answer and wondering where they went wrong, so you just might be right that the book is incorrect!

Hi everyone, I took a look at previous questions similar to this one and it seems like I am doing everything correctly, but I still get the wrong answer. Any advice to where I am going wrong would be greatly appreciated!Homework Statement A point charge (m=1.0g) at the end of an insulating...

GREAT! Thanks so much for your help and patience!:smile:

Mmm yes I think I multiplied by mg because I remembered doing that in previous problems...but now I realize those problems involved normal force which this one doesn't :bugeye: I'm doing this course by correspondence and I don't think I'm grasping some of these concepts as strongly as I should...

Ohhhh, thanks Doc Al, I think I'm understanding better now. I've found the tension, but that tension is acting on the bead from two sources. Full value of tension is found in AB, and this value added to the strictly horizontal part of the tension from BC would equal total force in the x...

Nice. Ok, so to find the horizontal force from the tension I went: tan53.1= 1.23/x x= 0.92 I'm not quite sure what you mean by no calculation being needed for BA. Is is the same value as the horizontal force from the tension? Would the equation FT=Ffrom a to b/2sinθ be applicable here...

Okay, that makes sense. So like you said, the horizontal tension from CB and AB will equal the net force, which equals maR. To find the horizontal tension in CB, I went mg(cosθ) = (9.8)(0.1)(cos53.1) to find horiztonal CB = 0.59. Now, to find AB I do the same thing of mg(cosθ) =...

Hi everyone. I've seen this question posted before so I apologize, but I have a hard time understanding the responses and some posts in different threads seem to go different directions (I don't know what to follow!). I've given the problem a try and would greatly appreciate knowing if I'm on...