If both bars are made of a good conductor, then their specific heat capacities must be different. If both are metals, specific heat capacities of different metals can vary by quite a bit, eg, both are in kJ/kgK, Potassium is 0.13, and Lithium is very high at 3.57 - both of these are quite good conductors.
If one of the bars is a good conductor and the other is a good insulator, then, after the surface application of heat, the temperatures at the surfaces are almost bound to be different. This is because the heat will be rapidly conducted into the body of the conducting bar, soon achieving a constant temperature throughout the bar. Whereas, with the insulator, the heat will tend to stay where it's put, heating the bar considerably over that area. As the heat slowly conducts into the bar, it will also start to cool from its surface, because it's so hot, and even if it has the same heat capacity as the other bar, which might be possible, it will eventually reach a lower, steady temperature throughout.
Answer:
56
Explanation:
I just want the points to be completely honest with you.
Straight
You already have to momentum of walking forward, and going back and forth are the same distance. If you go back then you would have to stop, turn and walk, but if you go forward you just have to walk.
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
