Answer:
Explanation:
Distance of 2 nd order fringe x₁ = 2 λ D/d [ λ is wave length , D is distance of screen , d is slit distance.
x₁ = 2 x 660 x10⁻⁹x 2.55/.65 x 10⁻³.= 5.17846 x 10⁻³ m.
Distance of fringe for 2nd radiation = 5.178 x 10⁻³ -1.17 x 10⁻³ = 4.008 x10⁻³
4.008 x 10⁻³ = 2 x λ x 2.55/.65 x 10⁻³
λ = 511 nm approx.
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For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:

The second equation is going to the total force acting on the charge q3:

k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:

Let's solve this for r_1^2:

Now we have a quadratic equation with following parameter:

We know that two solutions are:

We need a positive solution. When we plug in all the numbers we get:
Answer:
a. Liquid state
b. At point R. The physical state of water at its boiling point temperature of 100 degree Celsius will be both liquid state as well as gaseous state.
c. at 110 degrees it is in gaseous state
Velocity is speed plus the direction of the speed.