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kkurt [141]
3 years ago
8

You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen

ts
1. Shining the green laser through grating A you see the first maximum 1 meter away from the center
2. Shining the red laser through grating B, you see the first maximum 1 meter away from the center

In both cases, the gratings are the same distance from the screen.

(a) What can you deduce about the gratings?
(b) What would you observe if you shone the green laser through grating B:?

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
b. (a) grating B has more lines/mm; (b) the first maximum less than 1 meter away from the center
c.(a) grating B has more lines/mm; (b) the first maximum more than 1 meter away from the center
d. (a) grating B has more lines/mm: (b) the first maximum 1 meter away from the center
e. (a) grating A has more lines/mm; (b) the first maximum more than 1 meter away from the center
f. (a) grating A has more lines/mm; (b) the first maximum 1 meter away from the center
Physics
1 answer:
Nikolay [14]3 years ago
5 0

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

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Answer:

75joules

Explanation:

Workdone = force x distance

workdone = 25 x 3

workdone = 75joules

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Moving company uses a machine to raise a 900 Newton refrigerator to the second floor of a building machine consists of a single
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Mechanical advantage of a machine is the ratio of the output force over the input force or M=Fo/Fi. Since M=1, Fi=Fo, or the input force is equal to the output force. This means that to raise the refrigerator that weighs 900 N, we need the same input force of 900 N, or Fo=Fi=900 N. 
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A small sphere with mass 1.5g hangs by a thread between two parallel vertical plates 5cm apart. The plates are insulating and ha
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The tralational equilibrium condition allows finding that the electric potential is   V = 4.8 10¹¹ V

Given parameter

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  • The charge on the sphere q = 8.9 10-16 C
  • Plate spacing d = 5 cm = 5.00 10-2 m

To find

  • The potential difference

Newton's second law states that the force is proportional to the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it establishes the condition for the equilibrium of the bodies

          ∑ F = 0

Where the bold indicate vector and F is the force

To use this equation we must fix a reference system with respect to which to carry out the decomposition and measurements of the forces; let's fix a system with the horizontal x axis and the vertical y axis, in the attachment I could see a free body diagram.

x- axis

     Fe - Tₓ = 0

     Fe = Tₓ

y-axis

     T_y - W = 0

     W = T_y

     mg = T_y

The electric force is

      Fe = q E = q V / d

let's use trigonometry to decompose the stress

     cos 30 =  T_y / T

     sin 30 = Tₓ / T

      T_y = T cos 30

      Tₓ = T sin 30

We substitute

        q V / d = T sin 30

        mg = T cos 30

It's solve the system of equations

         \frac{q \ V}{d \ m g} = tan 30

         V = \frac{d \ mg }{ q}\ tan \  30

It's calculate

         V = \frac{5.00 \ 10^{-2} 1.5 \ 10^{-3} \ 9.8}{ 8.9 10^{-16} } \ tan \ 30

         V = 4.768 10¹¹ V

In conclusion, using the equilibrium condition, we could find that the electric potential is V = 4.8 10¹¹ V

Learn more about equilibrium condition here:

brainly.com/question/1967702

7 0
2 years ago
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
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Answer:

ω = 22.36 Hz

f = 3.56 Hz

T = 0.28 s.      

Explanation:

a) The angular frequency (ω), can be calculated using the following equation:

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{x*m}}  

<u>Where:</u>

<em>k: is the spring constant = F/x</em>

<em>m: is the mass of the particle = 0.500 kg</em>

<em>F: is the force applied = 7.50 N       </em>

<em>x: is the displacement = 3.00 cm = 0.03 m </em>            

\omega = \sqrt{\frac{7.50 N}{0.03 m*0.500 kg}} = 22.36 s^{-1} = 22.36 Hz    

Therefore, the angular frequency of the motion is 22.36 Hz.

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Hence, the frequency of the motion is 3.56 Hz.

c) The period (T) is equal to:

T = \frac{1}{f} = \frac{1}{3.56 Hz} = 0.28 s

Therefore, the period of the motion is 0.28 s.

I hope it helps you!

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v^2 -  u^2 = -2as

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a = (10)^2 - (20)^2/ - 2 * 60

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a = 2.5 m/s^2

At that time;

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a = 2.5 m/s^2

s = ?

Hence;

u^2 = 2as

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s = (20)^2/ 2 *  2.5

s = 400/5

s = 80 m

Hence, the bus will travel a further 20 m before coming to rest.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

7 0
2 years ago
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