Recent content by squirrelschaser

Homework Statement So the equipotential surface of a point charge is sphere with the charge in the center, and the equipotential surface of a infinite line is a cylinder with the line of charge as the axis. I was wondering what is the shape of the equipotential surface of a infinite plane...

The solutions I have showed the second answer as the correct answer. Really? That's the solution provided to me. Glad to know I wasn't paranoid or something. Is there any additional information that would make solving this question using kinematic possible, then?

Homework Statement Bob starts at rest from the top of a frictionless ramp. At the bottom of the ramp, he enters a frictionless circular loop. The total mass of the child and the cart he sits in his m. What must the height of the ramp be in order for the cart to successfully traverse the loop...

Interesting. I see that since the centrifugal force wouldn't be in the x or y direction in my coordinate system. Guess I'm just used to blindly resolving the force of gravity because of all those ramp questions. Much thanks for the insights.

That's exactly what I did. That's exactly what I did. Fy = 0 in my coordinate system but Fx != 0 since there will be mgcos(theta). This would mean that the mass will be sinking down into the cone and since it's stated that in the question it goes around a horizontal circle, I can't resolve the...

Whoops. Sorry for my incompetence. I added the question "What is the normal force exerted on the puck by the inner surface of the cone in terms of m, g, and theta?"

Homework Statement A hollow cone is put upside-down with its symmetry axis vertical. The surface of it makes an angle of theta with the vertical direction as shown in the figure . A small puck of mass m slides without friction on the inner side of this cone and remains within a horizontal plane...

-(3GMm/4R) -4GMm/4R = .5mV^2 - (3GMm/6R) - (4GMm/2R) -7GMm/4R = .5mV^2 - 15GMm/6R 9GMm/12R = .5V^2 9GM/12R = .5V^2 18GM/12R = V^2 sqrt(1.5GM/R) = V. Alrighty, I got the answer. I knew that PE isn't a vector, but it didn't make sense to me (logically) that a PE in another direction is...

?. Well maybe not arbitrarily, I thought since the 2 PE are in opposite direction, one would be positive and one would be negative. I will try this it with all the PE as negative values.

Not sure. I arbitrarily chose the attraction to the left as positive (PE from the moon) and the attraction to the right as negative (PE from the Earth). Should they all be negative?

4R. IT's halfway between the center and center point. So 8R/2.

Homework Statement Planet 1 has mass 3M and radius R, while Planet 2 has mass 4Mand radius 2R. They are separated by center-to-center distance 8R. A rock is placed halfway between their centers at point O. It is released from rest. Ignore any motion of the planets. The rock is released from...

I'm dumb. Much thanks.

Homework Statement Find the ## lim _{x-> -1+} sqrt(x^2-3x)-2/|x+1| ## Homework EquationsThe Attempt at a Solution I can only solve it using l'hopital rule and would like to know the steps of solving it without using it. ## lim _{x->-1+} (2x-3)/|1|= -5/4 ##