Recent content by Masschaos
-
M
What is the orbital period of a satellite in a low-Earth orbit?
Oh yes, I do see. That is almost exactly what is required. I don't suppose you'd know any reason for keeping it to the linear term? Perhaps because h << R the other terms become negligible. I think that is reasonable! Thank you very much. That was much simpler than I thought.- Masschaos
- Post #3
- Forum: Advanced Physics Homework Help
-
M
What is the orbital period of a satellite in a low-Earth orbit?
Homework Statement Consider a satellite in a circular, low-Earth orbit; that is, its elevation above the Earth’s surface is h ≪ R⊕. Show that the orbital period P for such a satellite is approximately P=C(1+ 3h/[2R⊕]). Homework Equations P2 = (4pi2)/(GM) * a^3. (G - gravitational...- Masschaos
- Thread
- Orbit Period
- Replies: 2
- Forum: Advanced Physics Homework Help
-
M
Buoyancy - Maximum mass a styrofoam sphere can hold in water.
Oh wow. Just a multitude of stupid mistakes! The diameter is 20cm so the radius is 10cm. AND the volume of a sphere is 4/3 pi r^3, I was doing the area of a circle. Argh, I'll hang my head in shame now.- Masschaos
- Post #3
- Forum: Introductory Physics Homework Help
-
M
Buoyancy - Maximum mass a styrofoam sphere can hold in water.
Homework Statement Styrofoam has a density of . What is the maximum mass that can hang without sinking from a 20.0 -diameter Styrofoam sphere in water? Assume the volume of the mass is negligible compared to that of the sphere. Homework Equations FB = ρfVg Fg = mg The Attempt at a...- Masschaos
- Thread
- Buoyancy Mass Maximum Sphere Water
- Replies: 3
- Forum: Introductory Physics Homework Help
-
M
How do you solve the integral ∫√(a² - x²) dx using substitution?
Oh of course! I can't believe I didn't see that. Thank you very much!- Masschaos
- Post #6
- Forum: Calculus and Beyond Homework Help
-
M
How do you solve the integral ∫√(a² - x²) dx using substitution?
So it is ∫a*cos(u) dx and not ∫a2 * cos2 (u) dx? I'm just struggling as I don't understand why some people are using ∫a2 * cos2 (u) dx. I've checked on wolfram Alpha also, and there is also the ∫a2 * cos2 (u) dx and I can't reconcile this from my above working.- Masschaos
- Post #3
- Forum: Calculus and Beyond Homework Help
-
M
How do you solve the integral ∫√(a² - x²) dx using substitution?
Homework Statement ∫√(a2 - x2) dx Homework Equations The Attempt at a Solution What I've understood so far is its a substitution integral. Through the instruction of my lecture notes, I have ∫√(a2 - x2) dx Let u = sin-1(x/a)...- Masschaos
- Thread
- Integral
- Replies: 5
- Forum: Calculus and Beyond Homework Help
-
M
Calculating Kp for Decomposition of NH4Cl at High Temperature
If, the activity of a solid is always 1. Then would Kp =[Products] Kp = [ 5.48*10-5] * [ 5.48*10-5]?- Masschaos
- Post #6
- Forum: Biology and Chemistry Homework Help
-
M
Calculating Kp for Decomposition of NH4Cl at High Temperature
Ah! So I am wrong. I assumed it was pressure at equilibrium. And after some searching I thought I had come up with the right answer. Thanks! I'll keep looking.- Masschaos
- Post #5
- Forum: Biology and Chemistry Homework Help
-
M
Calculating Kp for Decomposition of NH4Cl at High Temperature
Well, the thing that confused me is that NH4Cl is a solid, so it is not counted in the calculation for Kp. Doesn't that mean it will be [products]/0? Or do you just not divide the product by anything? What do you mean about the solid? it says its heated, and gas forms so presumably it gets...- Masschaos
- Post #3
- Forum: Biology and Chemistry Homework Help
-
M
Calculating Kp for Decomposition of NH4Cl at High Temperature
A 9g sample of solid NH4Cl is heated in a 4L container to 924.4oC and decomposes according to the following equation. NH4Cl(s) -> HCl(g) + NH3(g) at equalibrium the partial pressure of NH3 (g) is 1.346 atm. Calculate the equilibrium constant for Kp for this reaction Homework Equations...- Masschaos
- Thread
- Reaction
- Replies: 5
- Forum: Biology and Chemistry Homework Help
-
M
Location of a virtual image in a convexed mirror
So, just say that the above question was for a concave mirror instead of a convexed mirror, how would the equations differ?- Masschaos
- Post #16
- Forum: Introductory Physics Homework Help
-
M
Location of a virtual image in a convexed mirror
Ahh. I see, thanks. If it interests you, my two conflicting sources, which must have caused me some confusion were http://www.physics.uq.edu.au/people/mcintyre/vergences/optics/go_mag.html http://www.physicsclassroom.com/class/refln/u13l4d.cfm Thanks once more!- Masschaos
- Post #14
- Forum: Introductory Physics Homework Help
-
M
Location of a virtual image in a convexed mirror
Oh, Ok. Thanks. Although, this means the site I was given for this was wrong. It clearly states the opposite. However, it does make more sense this way. Thank you kindly.- Masschaos
- Post #12
- Forum: Introductory Physics Homework Help
-
M
Location of a virtual image in a convexed mirror
The magnification formula is m = v/u v = -0.084 u= 0.14 thus m = -.6 This implies that the image is inverted. However, my ray diagram begs to differ. I think I need a little more practices with ray diagrams. Edit: I found a source quoting that the m = -v/u If this is true, then all my problems...- Masschaos
- Post #10
- Forum: Introductory Physics Homework Help