Post Lab on Tension: Qs 1 & 2 Answered, Need Help with Q3

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The discussion focuses on verifying answers for a post lab on tension, specifically questions 1 and 2. The first answer confirms that the tensions are identical, making it impossible to distinguish between them. The second answer explains that transferring mass from cart B to cart A decreases the tension between the two carts. For question 3, it is noted that the tension remains consistent throughout the system when stationary, which allows for the calculation of the hanger's mass. Overall, the participants aim to clarify their understanding of tension in the context of the lab experiment.
Habibrobert
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Hello,

I am stuck on this post lab question. I have tried to answer questions 1 and 2 but I am unsure of my answers, could someone verify them to make sure I did this correctly? Also, I have no idea how to do number 3! I included screen shots of the problem.

Thanks very much.
 

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1. I agree with your answer - the tension are identical, so one cannot distinquish between the two.
2. When some of B's mass is transferred to A the tensions between the two carts will decrease, so I agree with you again - the pull is less in 2.
3. The tension in all the strings, or all of the tensions, are the same throughout the system when held still. Form this you can calculate the mass of the hanger.
 
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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