- #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
- 42,670
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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.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook.
Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water.
I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
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