Recent content by PeachBanana

For clarification: Was I incorrect in stating Δt initial was 60?

Oh, wow. Should I be thinking about γ = 1.25? Because...60 / 1.25 = 48.

I think after re-reading and thinking and google searching it has dawned on me, haha. So we know 60 beats per minute means one beat / one second which is the same as... 1 / 60 s.

I still don't think I quite understand...I think I might be stuck on the whole minutes to seconds thing. Edit: Does it even matter that it's in minutes since the final answer is in minutes too?

Period and frequency are inversely related. So since heart beat is frequency, 60 bpm, period would be 1/60 min. My next problem is the SI unit of time is not minutes but if I convert I get 1/3600 seconds.

Homework Statement A spaceship approaches Earth with a speed of 0.6c. A passenger in the spaceship measures his heartbeat as 60 beats per minute. What is his heartbeat rate according to an observer who is rest relative to Earth? 1. 48 beats per minute 2. 56 beats per minute 3. 65 beats per...

I see now. sin 45° * 12.8*10^6 N = 9.0 *10^6 N

Wow, it took me a little bit but now I understand the first part (the square root and the hypotenuse). Now I'm confused as to how he went from the vertical component of the force to the 9.0 * 10^6 N? Did he subtract the horizontal component so all he'd have left to compare are vertical components?

Does the square root come from the Pythagorean Theorem? Also, I don't understand what he did with the 12.8*10^6 N.

I should have just posted this initially. I hope the attachment helps.

The 0.075 m^2 came from me taking 0.15 m^2 / 2. I'm looking at the answer key now and what the instructor has is: F = (8.99 * 10^9 Nm^2/C^2)(0.004 C)^2 / 2(0.075 m^2) That's not the entire answer but that's the only part where I see 0.075 m.

Sorry, I was trying to mimic how it did on an exam. I see where I left charges - I will update with the correct attachment. The question in the exam didn't specify exact positions which I why I didn't. Relative to the +4 mC at the top of the circle, the two -4 mC will create a vector...

Homework Statement Two +4 mC charges and two -4mC charges are arranged on a circle of diameter "d" as shown in the figure below. The magnitude and direction of the force on the +4mC charge at the top of the circle is: a) 6.39*10^6 N radially inward b) 1.2 * 10^7 N radially inward c) 2.4...

Aha! 0 N. (right?)

Homework Statement Two point charges with charges of +5μC and -100μC are separated by 15 cm. The magnitude of the force between these charges is: a) 1.99 N b) 4.02 N c) 20 N d) 199 N e) 400 N Homework Equations F = q1q2 * k / r^2 The Attempt at a Solution F =...