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A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?

Natali [406]

Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.

The ruling separation is d=1/(470mm-1)

$= 2.1 \times 10 {}^{ - 3}mm$

Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.

Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.

we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is,find the greater integer value of m for which mλ<d.

since,d/λ

$= 538 \times 10 {}^{ - 9} m/2.1 \times 10 {}^{ - 6}$

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

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4 0

2 years ago

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

3 years ago

If it takes 100 N to get a 10 kg object to accelerate at 10m/s/s, how much force will it take to get a 20 kg object to accelerat

yan [13]

Answer: C) 200 N

Explanation:

The force $F$ is defined as:

$F=m.a$

Where:

$m=20 kg$ is the mass of the object

$a=10 m/s^{2}$ is the acceleration

Then:

$F=(20 kg)(10 m/s^{2})$

Finally:

$F=200 N$

Hence, the correct option is C.

7 0

3 years ago

Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per

Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 $^{\circ}$ to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0

3 years ago

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$