Answer:
It comes out the positive side of the battery and goes in to the negative side of the battery
Explanation:
There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.
In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.
However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative
Answer:
a) d₁ = 247.8 μm
d₂ = 205.3 μm
b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm
Explanation:
a)
The formula for Michelson Interferometer is derived to be:
d = mλ/2
where,
d = distance moved
m = no. of fringes
λ = wavelength of light
For JAN, we have following data
d = d₁
m = 818
λ = 606 nm = 606 x 10⁻⁹ m
Therefore,
d₁ = (818)(606 x 10⁻⁹ m)/2
<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>
For LINDA, we have following data
d = d₂
m = 818
λ = 502 nm = 502 x 10⁻⁹ m
Therefore,
d₂ = (818)(502 x 10⁻⁹ m)/2
<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>
b)
The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,
Resultant Displacement = Δd = d₁ - d₂
Δd = 247.8 μm - 205.3 μm
<u>Δd = 42.5 μm (in the direction of JAN)</u>
True, air moves from high to low pressure wanting to expand as much as possible.
