Recent content by cs44167

So the third charge would have to be to the left of the origin, correct? Then you would have negative-positive-negative. From there I’m struggling with Coulomb’s Law because we don’t know the magnitude of the charge on the third object. I have q3q1/x^2 =q1q2/3^2 but we don’t know x or q3.

I set the electrostatic force exerted by the object at (0,0) and (3,0) equal to each other, dividing out k and q2. I was left with q1/d^2 for both terms and substituted in the given charges for each object. I then replaced d^2 for the object at (0,0) with “x^2” and d^2 for the object at (3,0)...

Yes, the 1.14 was a typo - it should’ve been 1.44 which I had in my calculations. What does it mean for the force to act at the center? Would that divide the radius into halves?

I tried using r * f * sin theta and calculated this: 1.14 m * 9.80 m/s/s * 7.78 kg = 109.7 N*m this was wrong; I needed three significant figures so I did 1.10E2 N*m which was also wrong. Since the torque is due to gravity; would it be -1.10E2 N*m since it’s angle is -270° which is -1?

I know the weight acts on the pen directly down, and the friction acts on the pen up the slope of the incline. I also know the components of the weight, where the parallel component is parallel to the slope, and the perpendicular component in perpendicular to the incline and has the same...

I tried using coefficient of friction = friction / Normal force, but needed a value for friction. I then tried to find the friction using a = f/m, but was unsure of which value to plug in for force. Simply finding the force given a and m will not yield the correct answer; the net force must be a...

It’s on a website and gives a red x when the answer is incorrect. There’s never been an issue and the answer for acceleration is to be expressed in m/s/s. The only thing I was thinking was since acceleration is a vector if not having a negative sign was the issue. We have three submissions, the...

This question required measurements which are the following: my weight = 108 lbs crouching distance (the distance from my regular height to where I crouch) = 90.6 cm jump height = 60.4 cm I first converted lb to kg, and I got 49.09 kg. I then used this value for w = mg and inputted 9.80 for g...