Recent content by ova5676

Homework Statement What is the maximum order number possible for red light (630 nm) illuminating a double slit with separation 3.0 x 10^5 m. Texbook answer: 4.76 x 10^11 Homework Equations mλ = d*sinθ mλ = d*x/L The Attempt at a Solution So λ = 630 nm or 6.30 x 10^-7 m, and d =...

Homework Statement Earth is a satellite of the sun with an orbit radius of approximately 1.5 x 10^11 m. a) What is the Sun's mass? **Answer:** 2.0 x 10^30 kg b) If the Sun's radius is 6.96 x 10^8 m, how does the Sun's density compare with Earth's density? **Answer:** Psun = 1/4 Pearth...

I understand, but how would I get radial distance?

So if that's not the radius, then what do I use?

It's kind of late, so excuse my mind-numbness here but what do you mean by sum of the forces due to both X and Y? Do you mean Fnet = Fg between X and Y + Fg between Y and Z? or just Fg between X and Y? And for 57, r means radius, which is 300 km above the Earth's surface, isn't that the...

Homework Statement Answers: 56. 6.16 x 10^17 N, 57. 894 N Homework Equations Fg = G*m1*m2/r^2 g = G*M/r^2 The Attempt at a Solution 56. Since the closest planet to Z is Y I figured the gravitational force could be calculated by just finding the gravitational force between Z and Y...

Homework Statement In a lab, we essentially used an apparatus similar to this: In three experiments with several trials, we varied three things: centripetal force (the mass of the bottom), radius (of the circle by adjusting the alligator clip), and mass (rubber stopper). The trials were...

Anyone got anything?

Ah, I see what you mean. Hope you don't mind answering one last question please. If you look @ this situation: Basically an object moves from that one inclined plane with friction across a frictionless surface then up a frictionless plane. I'm not asking about calculating anything, but I'm...

Well in my specific case, they are at rest initially (three people are holding them) and then all three let go BUT one of them doesn't move (the one with the heaviest weight, by the way). So, would I assume the first mass and the last mass have the same acceleration but the middle mass would...

So acceleration is the same for each mass, is that what you mean?

So in other words, that would be: First mass = T1 - Fg1 Second mass = Fg2 - 2(T1) Third mass = T1 - Fg3 ?

I think if you just listed the forces involved in the x- and y- components and you explain why the component section with forces perpendicular to the Tension should be used, we would all be on the same page.

I was wondering about the two strings thing, but I was told that if pulleys were held at a ceiling instead of an inclined plane like this thread, that it's just one string? How positive are you that it's two strings? If it's two strings, two tensions so: First mass = T1 - Fg1 - same thing...

So it's Fgy, right? If you're looking at the diagram as a rotated coordinate system, Tension would be in the horizontal and Fg would have x- and y- components, correct?