How Do Relevant Equations Help Solve Problems?

Thread starter postmalone
Start date Jul 13, 2022

AI Thread Summary

Relevant equations play a crucial role in problem-solving by providing a framework for understanding the relationships between different variables. They guide the approach to a problem, allowing for systematic analysis and application of mathematical principles. Demonstrating reasonable effort before seeking help is essential, as it shows engagement with the material and helps clarify the specific challenges faced. By articulating one's thought process and the equations considered, individuals can receive more targeted assistance. Ultimately, utilizing relevant equations effectively enhances problem-solving capabilities.

Jul 13, 2022

postmalone

New poster has been reminded to always show their work when starting schoolwork threads

Homework Statement: An interesting problem using capacitors and moving charges in a magnetic field

Relevant Equations: r=mv/qb

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Jul 13, 2022

kuruman

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According to PF rules, you need to show reasonable effort before receiving help. Tell us what you think, how you would approach the problem and what is the role of the "relevant equations" in the solution.

Nevertheless, the answer to the question posed in the title is "Yes, we can solve this problem."

Thread 'I've come across another fluid pressure problem I don't understand'

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?

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Thread 'Variable mass system : water sprayed into a moving container'

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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Thread 'Understanding Forces and their Vector Components'

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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