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ale4655 [162]
3 years ago
8

When Casey looks at a CD or DVD she notices that there is a rainbow pattern that is created by the light. She became curious and

shone light directly at it and a pattern was created on the wall. What is the cause of this pattern?

Physics
2 answers:
ipn [44]3 years ago
6 0

Answer:

Explanation:

diffraction

LekaFEV [45]3 years ago
4 0

Answer:

Interference

Explanation:

A CD is made up of thousands of pits arranged in the form of spiral tracks.These repeated patterns of pits form a thin film.

This film has the thickness of the order of the wavelength of the light which causes the interference of the waves of the light waves. The incident ray of light gets reflected from the top layer of the film partially and then after striking the bottom layer of the film creating a phase difference responsible for the constructive and destructive interference.

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Lapatulllka [165]
Pop rocks and soda..... baking soda and vinegar ...
4 0
3 years ago
Read 2 more answers
What is the kinetic energy of a 36N toy car which is moving at 5 m/s?
vagabundo [1.1K]
Correct option is
C
y
max
​
=11 m
Work done =mgh

2
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×base×height=mgh
(area under the curve is work done)

2
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​
×11×100=5×10×h

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3 0
3 years ago
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha
lbvjy [14]

Answer:

Electric flux;

Φ = 30.095 × 10⁴ N.m²/C

Explanation:

We are given;

Charge on plate; q = 17 µC = 17 × 10^(-6) C

Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²

Angle between the normal of the area and electric field; θ = 4°

Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m

Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²

The charge density on the plate is given by the formula;

σ = q/A_p

Thus;

σ = (17 × 10^(-6))/(180 × 10^(-4))

σ = 0.944 × 10^(-3) C/m²

Also, the electric field is given by the formula;

E = σ/ε_o

E = (0.944 × 10^(-3))/(8.85 × 10^(-12))

E = 1.067 × 10^(8) N/C

Now, the formula for electric flux for uniform electric field is given as;

Φ = EAcos θ

Where A = πr² = π × 0.03² = 9π × 10^(-4) m²

Thus;

Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4

Φ = 30.095 × 10⁴ N.m²/C

3 0
3 years ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0
3 years ago
A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is oriented
oee [108]

Answer:

Magnetic flux, \phi=2.22\times 10^{-3}\ Wb

Explanation:

It is given that,

Magnitude of the magnetic field, B = 0.078 T

Radius of circular loop, r = 0.1 m

The field is oriented at an angle of θ = 25° with respect to the normal to the surface. The magnetic flux through the surface is given by :

\phi=BA\ cos\theta

\phi=0.078\times \pi \times (0.1)^2\ cos(25)

\phi=0.00222\ Wb

or

\phi=2.22\times 10^{-3}\ Wb

So, the magnitude of magnetic flux through the surface is 2.22\times 10^{-3}\ Wb. Hence, this is the required solution.

4 0
4 years ago
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