- #1
aChordate
- 76
- 0
Homework Statement
Homework Equations
B=F/|q0|vsinθ
V=IR
The Attempt at a Solution
V=500m/h = 8.05m/h = 223m/s
L=35m
B=0.50gauss
0.50g=F/|q0|*223m/s*sin90
Voltage difference of the wing of an airplane in Earth's magnetic fiel
- Thread starter aChordate
- Start date
AI Thread Summary
The discussion centers on calculating the voltage difference across an airplane wing in Earth's magnetic field. Key equations mentioned include B=F/|q0|vsinθ and V=IR, with a focus on the conversion of gauss to tesla for accurate calculations. The velocity of the airplane is given as 223 m/s, and the magnetic field strength is noted as 0.50 gauss. Participants suggest that the equation V = vBL sin θ may be more applicable to the problem. Overall, the conversation emphasizes the need for precise unit conversions and relevant equations in solving the problem.
B=F/|q0|vsinθ
V=IR
V=500m/h = 8.05m/h = 223m/s
L=35m
B=0.50gauss
0.50g=F/|q0|*223m/s*sin90
- #2
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Wouldn't V = vBL sin θ be more to the point?aChordate said:
I guess you mean v=500mph = 805kph = 223m/s
- #3
rude man
Science Advisor
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aChordate said:
You need to change gauss to tesla.
Also, none of your proffered equations are directly relevant. Consider haruspex's suggestion.
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question.
Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
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