Ohm's Law: Examining a Sample's Resistance

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The sample exhibits different resistances at varying currents, indicating it does not obey Ohm's Law consistently. Calculations show a resistance of 0.5 ohms at 2V with a current of 4A and 0.4 ohms at 4V with a current of 10A, suggesting non-linear behavior. The book's answer of 0.5 ohms at 1V is questioned, as it does not align with the calculated values. Participants suggest plotting the resistance against current for clarity, but the varying resistances complicate a straightforward interpretation. The discussion highlights confusion over the problem's accuracy and the nature of the sample's resistance.
blackout85
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A certain sample carries a current of 4 A when the potential difference is 2V and a current of 10 A when the potential difference is 4V. This sample:
A) obeys Ohm law
B) has a resistance of 0.5 ohms at 1V
C) has a resistance of 2.5 ohms at 1V
D) has a resistance of 2.5 ohms at 2V
E) does not have a resistance

my work:
R= V/I= (2/4) = .5
R= (4/10)= .4

The answer according to the book the answer is B. I do not see how they got that answer could someone please explain. Thank you
 
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Maybe try plotting R(I) to see if something pops out. You have two data points already -- do any of the answers A-E also fit with the other two data points?
 
Is there a way to go about it mathematically without graphing?
 
I suppose, but even with graphing the two points (and the unknown point at 1V), it's not obvious to me how to proceed. The resistances are different at each of the points, and are not equal to the slope of the line between them. Even if the resistance is varying linearly with current, you will not get the same answer of 0.5 Ohms at both 1V and 2V. Are you sure the problem and answer are stated correctly?
 
Thread 'Variable mass system : water sprayed into a moving container'

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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