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kakasveta [241]
3 years ago
9

If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes d

isappear? Assume the fringes are viewed by a detector sensitive to both wavelengths.
Physics
1 answer:
omeli [17]3 years ago
4 0

Answer:The central spot becomes Dark

Explanation: it become dark because as the wavelength reduces,the velocity in the detector decreases, this time by 90degrees

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3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
4 years ago
A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk
bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

5 0
3 years ago
If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?
aivan3 [116]

Initial velocity = \(v_0\)

acceleration in the downward direction = -9.8 \(\frac {m}{s^2}\)

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

\(v^2 = {v_0}^{2} + 2aH

\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410

\({v_0}^{2} = 2 \times 9.8 \times 0.410\)

\(v_0 = 2.834 \frac {m}{s}\)

Speed with which the flea jumps = \(2.834 \frac {m}{s}\)

4 0
3 years ago
True or false, russian twists focus to strengthen the latissimus dorsi
Oliga [24]
The answer is true hope that helped!!
5 0
3 years ago
Read 2 more answers
What is the weight of a<br> 63.7 kg person?
IRINA_888 [86]

Answer:

A person that weighs 63.7 kg weighs 63.7 kg

Explanation:

Too complicated to explain.

6 0
3 years ago
Read 2 more answers
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