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jek_recluse [69]
3 years ago
13

Light from a helium-neon laser (λ = 633 nm) illuminates two slits spaced 0.50 mm apart. A viewing screen is 2.5 m behind the sli

ts. What is the spacing between two adjacent bright fringes?
Engineering
1 answer:
serg [7]3 years ago
7 0

Answer:

Explanation:

Given

wavelength \lambda =633\ nm

distance between two slits d=0.5\ mm

Screen is placed at a distance L=2.5\ m

Location of a (n+1)th bright fringe is given by

x_{n+1}=\frac{L}{2d}(n+1)\lambda

for nth bright fringe

x_n=\frac{L}{2d}(n)\lambda

Distance between two bright fringes

x_{n+1}-x_n=\frac{L}{2d}\cdot \lambda

x_{n+1}-x_n=\frac{2.5}{2\times 0.5\times 10^{-3}}\cdot 633\times 10^{-3}

x_{n+1}-x_n=1.582\ mm

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A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul
Ghella [55]

Answer:

a) 1.16 m/s

b)  1/216000

c)  (√15)/6480000

Explanation:

The parameters given are;

Length of boat prototype, lp = 30 m

Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)

vm = 9 × (1/60)^(1/2) = 1.16 m/s

b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

Fm/Fp = (1/60)³ = 1/216000

c) The ratio of the model to prototype power  pm/p_p = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power  pm/p_p = √(1/60) × (1/60)³

pm/p_p = √(1/60) × (1/60)³ = (√15)/6480000

6 0
3 years ago
What is Elon Musk mad about?
boyakko [2]

Answer:

Tesla CEO Elon Musk tweeted that the company's stock was too high, and it immediately dropped in value. The tweet may have violated a deal Musk made with the SEC about his tweets and Tesla. Musk also tweeted patriotic lyrics, said his girlfriend Grimes is mad at him, and noted that their child is due on Monday

6 0
3 years ago
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  h_1=\dfrac{\Delta h}{\Delta T}

  h_1=\dfrac{2406}{20}

h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

 h_2=\dfrac{total\ heat}{\Delta T}

 h_2=\dfrac{2,544}{20}

h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

             =\dfrac{127.2-120.3}{120.3}\times 100

                   =5.75 %

So Percentage change 5.75 %.

3 0
2 years ago
An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa
mote1985 [20]

Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

4 0
3 years ago
The modifications of superheat and reheat for a vapor power plant are specifically better for the operation which of the followi
MatroZZZ [7]

The modifications of superheating and reheat for a vapor power plant are specifically better for the operation which of the following components b.Boiler.

<h3>What are the primary additives in the vapour strength cycle?</h3>

There are 5 steam strength cycles: The Carnot cycle, the easy Rankine cycle, the Rankine superheat cycle, the Rankine reheat cycle and the regenerative cycle.

  1. Central to expertise the operation of steam propulsion is the primary steam cycle, a method wherein we generate steam in a boiler, increase the steam via a turbine to extract work, condense the steam into water, and sooner or later feed the water again to the boiler.
  2. Reheat now no longer best correctly decreased the penalty of the latent warmness of vaporization in steam discharged from the low-stress quit of the turbine cycle, however, it additionally advanced the first-rate of the steam on the low-stress quit of the mills via way of means of decreasing condensation and the formation of water droplets inside the turbine.

Read more about the Boiler:

brainly.com/question/17362931

#SPJ1

7 0
2 years ago
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