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Alekssandra [29.7K]
3 years ago
10

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What

is the wavelength λ of light that creates a first-order fringe at 22.0°?
Physics
1 answer:
Svetlanka [38]3 years ago
6 0

Explanation:

It is given that,

Wavelength of red laser light, \lambda=632.8\ nm=632.8\times 10^{-9}\ m

The second order fringe is formed at an angle of, \theta=53.2^{\circ}

For diffraction grating,

d\ sin\theta=n\lambda

d=\dfrac{n\lambda}{sin\theta}, n = 2

d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}

d=1.58\times 10^{-6}\ m

The wavelength λ of light that creates a first-order fringe at 22 is given by :

\lambda=d\ sin\theta

\lambda=1.58\times 10^{-6}\ sin(22)

\lambda=5.91\times 10^{-7}\ m

\lambda=591\ nm

Hence, this is the required solution.

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3 0
3 years ago
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A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is
GREYUIT [131]

Answer:

\omega = 22.67 rad/s

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

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now we know that work done is product of force and displacement

so here we have

W = F.d

W = (44 N)(0.9 m) = 39.6 J

now for moment of inertia of the disc we will have

I = \frac{1}{2}mR^2

I = \frac{1}{2}(7 kg)(0.21^2)

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now from above equation we will have

39.6 = \frac{1}{2}(0.154)\omega^2

\omega = 22.67 rad/s

5 0
3 years ago
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An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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7 0
2 years ago
Question:
exis [7]

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE      (PE = 0)

ME = KE

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ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE      (KE = 0)

ME = PE

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Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² =  m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s²  = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

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