In order for two objects to have the same temperature

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For two objects to have the same temperature, they must be in thermal equilibrium. Thermal contact alone is insufficient due to factors like radiation and material conductivity, which can affect perceived temperature. Options suggesting that objects share qualities without thermal equilibrium are incorrect. The consensus is that option e, indicating thermal equilibrium, is the correct answer. Therefore, thermal equilibrium is essential for two objects to have the same temperature.
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1. In order for two objects to have the same temperature, they must _____


a- be in thermal contact with each other
b-have all these qualities, except for thermal equilibrium
c- have all of these properties
d- have the same relative "hotness" or "coldness" when touched
e- be in thermal equilibrium




2. My guess is e but want to be sure.



3. a is not correct because of radiation i guess b and c don't sound right and for d some metals can be same temperature but because of conductivity one might feel hotter or colder
 
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I've no Idea what the qualities or properties mentioned in answers b and c are.
a is wrong, because of the existence of thermometers, melting ice, radiative equilibrium and also dumb luck. d is also wrong for the reason you mentioned, and e is correct.
 
i think so too b and c don't make sense:S a and d are definetaly wrong so e is the choice to go then for sure right?
 
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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