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Sparkle2009
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Homework Statement
In a youngs double slit experiment,if the slit widths are in the ratio 1:2,the ratio of intensities at maxima and minima will be ?
Youngs Double Slit: Intensities at Maxima & Minima
- Thread starter Sparkle2009
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- Double slit Slit
AI Thread Summary
In a Young's double slit experiment with slit widths in the ratio of 1:2, the ratio of intensities at maxima and minima is determined to be 1:2. The discussion includes attempts to solve the problem and references to relevant equations and images for clarity. Participants express frustration over the lack of responses to the question. The correct answer has been confirmed as 1:2, indicating a clear understanding of the intensity distribution in this setup. This highlights the relationship between slit width and intensity in interference patterns.
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?
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}...
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
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