<u>Answer</u>
A. Metals A and metals B
<u>Explanation</u>
Heat transfer takes place whenever there is temperature difference. When two bodies of different temperatures are brought together, heat energy will move from one body to the other until equilibrium temperature is reached.
In our case, heat transfer will take place in all four metals.
Metal A will transfer heat to the water since it's temperature is higher than that of water.
Metal B will also transfer heat to the water since it's temperature is higher than that of water.
Metal C will get heat from the water since it's colder than the water.
Metal D will also get heat from the water since it is colder than water.
Answer:
Simple awnser Do it yourself I really would help but I have no clue! Sorry
Explanation:
Answer:
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Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J