Hey all, I think that I'm doing this problem correctly but I'm getting an answer that's a couple thousand Kelvins off. Sorry if I should have posted this in the "advanced" section.
1. Homework Statement
Part of a flat region of Pluto has the Sun directly overhead. Its surface temperature is 58...
Homework Statement
A pocket of air rises 1000 meters. Estimate how much it cools. Use the atmospheric law to determine pressure at 1000 meters.
Homework Equations
P = P0 * e-Mgz/RT
PVγ = constant
TVγ-1 = constant
The Attempt at a Solution
Using the atmospheric law, I found the pressure at...
Would dV/dt just be the derivative of the right side of my equation above? That is, dV/dt = -(V0/RC)*e-t/RC?
Then, since I = C*dV/dt,
I = -3 volts/10 ohms * e(10 s)/((10 ohms)*C)
I feel like that doesn't help, because then I just have another unknown (I).
So I can arbitrarily say that Vc(10 s) is, say, 1.5 volts? And then solve for C, which would give 1.4 farads. Isn't that considered a really really high capacitance?
Homework Statement
Imagine that you have a light-bulb that has a resistance of about 10 ohms and that can tolerate a maximum voltage of 3 volts. Imagine that you want to connect this to a charged capacitor large enough to keep the bulb glowing reasonably brightly for more than 10 seconds...
Homework Statement
I have two equations. The first is for all of the forces on a hanging mass from a pulley. The second is for the sum of the torques about the pulley from which the mass hangs. I simply have to combine the equations to find the acceleration of the object. I have attempted every...
Homework Statement
So this is a small part of a much larger problem that I'm working on that I don't want to post here. Basically, I want to find the contribution to the torque about the hinge of a ladder BY the mass of the ladder itself. A ladder is propped open at an angle α. What is the...
Ok, yeah, I realize that that's not the equation and it ought to be α.
I decided to try a different approach. The torque on the pulley is equal to rg(m_2 - m_1).
The torque is also equal to Iα, or I(a/r)
Combining my two equations for the torque, I get
rg(m_2 - m_1) = (2/3)(m_pulley)r^2 *...
OK, I've tried this and here's what I've got (T_2 is the tension in the string connecting m_2, T_1 is the tension in the string connecting m_1)
τ=T_2 * r - T_1 * r
a = αr
a will be the same as the acceleration of mass 2, because mass 2 is on a string connected to the pulley.
Iω = τ
ω = τ/I...
Homework Statement
Two blocks, of masses m1 and m2 (with m1<m2) are attached by a massless string of fixed length that runs over a pulley which is free to rotate about its fixed center. The pulley has mass m3, radius r3, and rotational moment of inertia I = (2/3) m3 r3^2 about its center (it...
Just because I think this is intrinsically related: a hand pulls a block connected to a spring away from a wall to which the other end of the spring is attached. The initial and final velocity of the ball is zero. Kinetic energy doesn't change as a result. Is the work then zero?
I feel like I'm...