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Sindrei [870]
4 years ago
13

Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang

le of 40.0° from central maximum. The same slit, when illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum. What is the wavelength of this new light?a. 256 nm b. 283 nm c. 307 nm d. 329 nm e. none of these
Physics
1 answer:
VLD [36.1K]4 years ago
5 0

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe =  λ / d ,   λ is wavelength and  d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second  dark fringe =  2 x λ / d ,   λ is wavelength and  d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 =  2 x λ / 410

λ = 307.5 nm.

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A submarine periscope uses two totally reflecting 45-45-90 prisms with total internal reflection on the sides adjacent to the 45
Likurg_2 [28]

Answer

Given,

Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.

refractive index of water, n_a = 1.33

refractive index of glass, n_g = 1.52

When the light enters the water, water will act as a lens and when we see the object from the periscope the object shown is farther than the usual distance.

7 0
3 years ago
A Tennis ball falls from a height 40m above the ground the ball rebounds
worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

4 0
3 years ago
Is this right at all
Alja [10]
It is correct! good job :)
7 0
3 years ago
Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109
Oksana_A [137]

Answer:

 A) F = 1.09 10 5 N, b) Yes  

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

               ρ = m / V

               .m = ρ V = ρ 4/3 π r³

The radius is half the diameter

               r = 1.9 10⁻² / 2 = 0.95 10⁻² m

               ρ = 8960 kg / m3

               m = 8960 4/3 π (0.95 10⁻²)³

               m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

             #_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

             #_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

               q = e / 10⁹    #_atom

               q = e / 10⁹    3,049 1023

               q = 3,049 10⁴  (-1.6 10⁻¹⁹)

               q = -4,878 10-5 C

Electric force is

             F = k q₁q₂ / r²

             F = k q² / r²

             

Let's calculate

             F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

              F = 1.09 10 5 N

This is a force of repulsion.

Part B

 The magnitude of this force is  in very easy to detect

4 0
3 years ago
(6.MS-ETS2-1(MA).) The electrons in __________ move about more freely than the electrons in insulators which is why this type of
Tresset [83]

Answer:

A) conductors

Explanation:

A conductor can be defined as any material or object that allows the free flow of current or electrons (charge) in one or more directions in an electrical circuit. Some examples of a conductor are metals, tungsten, copper, aluminum, iron, graphite, etc.

Basically, the main purpose of a conductor in physics is to provide a low-resistance path between electrical circuits or components. This low-resistance path is to ensure that the electrical components allows the free flow of electrons and thus, enabling charge transfer.

Hence, the electrons in conductors move about more freely than the electrons in insulators which is why this type of material can be used to create electric circuits because it would significantly provide a low-resistance path between the electric circuits.

8 0
3 years ago
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