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Alika [10]
3 years ago
5

Consider the following statements. A. Heat flows from an object at higher temperature to an object at lower temperature; B. Heat

flows from an object with higher thermal energy to one with lower thermal energy; C. Heat flows from an object in liquid state to an object in solid state. Which statements are true?
Physics
1 answer:
Monica [59]3 years ago
8 0

Answer:

A. Heat flows from an object at higher temperature to an object at lower temperature

Explanation:

The option A obeys the 2nd law of thermodynamics. The heat will flow from the object at higher temperature to the object at Lower temperature till they reach an equilibrial state.

Heat doesn’t necessarily flow from an object with higher thermal energy to an object with lower thermal energy because an object has a higher thermal energy when it’s mass is more than the other. This makes B wrong.

C is wrong because heat moves from an object with higher temperature to objects with Lower temperature regardless of the state of matter.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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Sort the following forces as relevant or not relevant to this situation. The symbols are defined as follows: normal force = n⃗ ,
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