1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia.

In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi

Ronch [10]

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference peak is given by the following expression

Width of each of bright fringe = λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁) / λ D/ d₂ = 13

then ( λ D/ d₁) / λ D/ d₂ = 13 / 2

= 6.5

no of fringes contained within each secondary diffraction peak = 6.5

6 0

4 years ago

As food passes through the alimentary canal, the presence of chemical secretions will occur in which order?

Natalija [7]

The first one....................

7 0

3 years ago

Read 2 more answers

A woman exerts a constant horizontal force on a large box. As a result, the box moves across a horizontal floor at a constant sp

d1i1m1o1n [39]

Answer:C

Explanation:

When a constant horizontal force is applied to the box, box started moving in the horizontal direction such that it moves with constant velocity $v_0$

Constant velocity implies that net force on the box is zero

i.e. there must be an opposing force which is equal to the applied force and friction force can serve that purpose.

So option c is the correct choice.

6 0

4 years ago

Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$