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MakcuM [25]
3 years ago
6

Monochromatic light falls on two very narrow slits 0.046 mm apart. Successive fringes on a screen 6.20 m away are 8.9 cm apart n

ear the center of the pattern. Part A Determine the wavelength of the light.
Physics
1 answer:
Elanso [62]3 years ago
8 0

Answer:

λ = 6.602 x 10^(-7) m

Explanation:

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is given as ;

y = mλD/d

Where;

D is the distance of the screen from the slits = 6.2 m

d is the distance between the two slits = 0.046 mm = 0.046 x 10^(-3) m

The fringes on the screen are 8.9 cm = 0.089 m apart from each other, this means that the first maximum (m=1) is located at y = 0.089 m from the center of the pattern.

Therefore, from the previous formula we can find the wavelength of the light:

y = mλD/d

So, λ = dy/mD

Thus,

λ = (0.046 x 10^(-3) x 0.089)/(1 x 6.2)

λ = 6.602 x 10^(-7) m

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An object of mass 3.00kg, moving with an initial velocity of 5.00 i^ m/s , collides with and sticks to an object of mass 2.00 kg
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The final velocity of the composite object is 6/5 m/sec.

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4 0
1 year ago
Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.
Triss [41]

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

                              ρA₁V₁ = ρA₂V₂

                                 A₁V₁ = A₂V₂

making V₂ the subject of the equation;

                               V_{2} = \frac{A_{1}V_{1}}{A_{2}}

the pipe is widened to twice its original radius,

                                r₂ = 2r₁          

then the cross-sectional area A₂ = 4A₁

                           ⇒  V_{2}= \frac{A_{1}V_{1}}{4A_{1}}

                                  V_{2}= \frac{V_{1}}{4}

This implies that the water speed will drop by a factor of  \frac{1}{4} because of the increase the pipe cross-sectional area.  

 The Bernoulli Equation;

     Energy per unit volume before = Energy per unit volume after    

        p₁ + \frac{1}{2}ρV₁²  + ρgh₁ = p₂ + \frac{1}{2}ρV₂²  + ρgh₂  

Total pressure is constant and P_{T} = P = \frac{1}{2}ρV₂²ρV²  

        p₁ + \frac{1}{2}ρV₁²  = p₂ + \frac{1}{2}ρV₂²

Making p₂ the subject of the equation above;

        p₂ = p₁ + \frac{1}{2}ρV₁² - \frac{1}{2}ρV₂²

But V_{2}  = \frac{V_{1}}{4} so,

        p₂ = p₁ + \frac{1}{2}ρV₁² - \frac{1}{2}ρ\frac{V_{1}^{2}}{4^{2}}      

       p₂ = 3.0 x 10⁴ + (\frac{1}{2} × 1000 × 12²) - ( \frac{1}{2} × 1000 × 12²/4² )

      P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³    

       P₂ = 9.79 × 10⁴Pa      

      P₂ = 9.8 × 10⁴Pa                      

4 0
2 years ago
a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoi
Free_Kalibri [48]

Answer:

Q₂ = 5833.33 J

Explanation:

First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:

η = W/Q₁

where,

η = efficiency of engine = 30% = 0.3

W = Work done by engine = 2500 J

Q₁ = Heat supplied to the engine = ?

Therefore,

0.3 = 2500 J/Q₁

Q₁ = 2500 J/0.3

Q₁ = 8333.33 J

Now, we find the heat discharged to lower temperature reservoir by using the formula of work:

W = Q₁ - Q₂

Q₂ = Q₁ - W

where,

Q₂ = Heat discharged to the lower temperature reservoir = ?

Therefore,

Q₂ = 8333.33 J - 2500 J

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7 0
3 years ago
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Time = distance divided by speed
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7 0
3 years ago
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