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weqwewe [10]
3 years ago
13

For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin

ter (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 7.9 cm wide.What is the width of the slit? answer in mm.
Physics
1 answer:
melomori [17]3 years ago
8 0

Answer:

a=0.0947mm

Explanation:

Given data

λ=680nm

L=5.5m

ω=7.9cm=0.079m

To find

Width of slit a

Solution

As the single slit of width a has a bright central maximum of width:

ω=2λL/a

a=2λL/ω

Substitute the given values

So

a=\frac{2*680*10^{-9}m*5.5m}{0.079m} \\a=9.47*10^{-5}m\\or\\a=0.0947mm

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Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
4 years ago
A train travels 90 kilometers in 1 hour, and then 82 kilometers in 5 hours. What is it’s average speed?
PIT_PIT [208]
Distance is 90km,time 1 hr
distance 2 is 82km,time 5 hrs
average speed=total distance travelled/total time taken.
90+82=172/5+1=6
average speed=28.7km over hour
4 0
3 years ago
A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
4 years ago
A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it mai
Sloan [31]

Answer: Hello the missing piece of your question is attached

question : Determine mass of steam that has entered ( in kg )

answer : 0.206 kg

Explanation:

V1 = 0.1 m^3 ,

v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg

V2 = 0.2 m^3

using the steam tables

at ; P = 1000 kPa, v' = 0.167 m^3/kg

U1 = 2321  KJ/kg

at ; P = 1000 kPa , T2 = 280°C

v'2= 0.2481 m^3kg

U2 = 2760.6

at ; P = 5MPa ,  T = 500°C

h1 = 3434.7 KJ/Kg

calculate final mass ( m2 )

M2 = V2 / v'2

      = 0.2 / 0.2481 =  0.806 kg

therefore the mass added =  m2 - m1

                                            = 0.806 - 0.6 =  0.206 kg

3 0
3 years ago
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