Answer:
R = 5.20 Ω, L = 6.24 10⁻⁴ H
Explanation:
The current in an RL circuit is
I = 
τ = L / R
In the problem they indicate the value of the voltage, the current and the time constant, for which the resistance must be found
The stable current is when enough time has passed (t »τ) after closing the circuit, therefore the exponential term is very small and we can neglect it.
I = E / R
R = E / I
let's calculate
R = 13.0 / 2.50
R = 5.20 Ω
now with this value we can find the inductance of the coil
τ = L / R
L = τ R
L = 1.20 10⁻⁴ 5.2
L = 6.24 10⁻⁴ H
Answer:As the size of a star increases, luminosity increases. If you think about it, a larger star has more surface area. That increased surface area allows more light and energy to be given off. Temperature also affects a star's luminosity.
Answer:
W = 30 J
Explanation:
given,
Work done = 10 J
Stretch of spring, x = 0.1 m
We know,
dW = F .dx
we know, F = k x


![W = k[\dfrac{x^2}{2}]_0^{0.1}](https://tex.z-dn.net/?f=W%20%3D%20k%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_0%5E%7B0.1%7D)

k = 2000
now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

![W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx](https://tex.z-dn.net/?f=W%20%3D%202000%20%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_%7B0.1%7D%5E%7B0.2%7D%20dx)
W = 1000 x 0.03
W = 30 J
Hence, work done is equal to 30 J.
Answer:
Metals, nonmetals and metalloids.
Explanation: