It's just asking you to sit down and COUNT the little squares in each sector.
It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it. That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.
(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)
Answer:
1) current = I
2) Resistance = V/I
3) current = 2I
4) resistance = V/2I
5) current = 3I
6) Resistance = V/3I
7) Current = 4I
8) Resistance = V/4I
Explanation:
When one bulb is connected across the battery then let say the current is given as I
Then resistance is given as

When two bulbs are in parallel with the battery then
total current becomes twice of initial current
so we have
current = 2I
Resistance of the circuit is now

When three bulbs are in parallel with the battery then
total current becomes three times of initial current
so we have
current = 3I
Resistance of the circuit is now

When four bulbs are in parallel with the battery then
total current becomes four times of initial current
so we have
current = 4I
Resistance of the circuit is now

Dipper effect of an oncoming train get louder as it approaches and sound diminishes as it goes away sound traveling
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