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HACTEHA [7]
3 years ago
6

Light of wavelength 500nm shines on a pair of slits. This produces an interference pattern on a screen 2.00m behind the slits. M

easuring from the center of the pattern, we find that the fifth dark spot is 4.00cm away. What is the spacing between the slits
Physics
1 answer:
grandymaker [24]3 years ago
7 0

Answer:

The  value is  d = 0.000125 \  m

Explanation:

From the question we are told that

   The  wavelength is  \lambda  =  500 \  nm =  500 *10^{-9 } \  m

    The  distance is of  the screen is  D =  2.0 \ m

    The  width of the fifth dark spot is y  =  4.00 \  cm  =  0.04 \  m

Generally the width of a fringe is mathematically represented as

           y =  \frac{ n  *  \lambda  *  D }{d }

=>        0.04  =  \frac{ 5  * 500 *10^{-9}  *  2 }{d}

=>        d =  \frac{ 5  * 500 *10^{-9}  *  2 }{0.04}

=>        d = 0.000125 \  m

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iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
A group of tissues working together to perform a similar function.
inn [45]

Answer:

organ

Explanation:

7 0
3 years ago
Read 2 more answers
The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?
Svetradugi [14.3K]

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

3 0
3 years ago
Calculate two stars appear to you to have the same brightness. if star a is 7.00 light-years away from you, while star b is 15.0
vagabundo [1.1K]

It  is eight times more than the star A.

<h3>What is luminosity  and on which it depends?</h3>

The luminosity of an object is a measure of its intrinsic brightness and is defined as the amount of energy the object emits in a fixed time.

luminousity depends upon the two factors are:

1) The star's actual brightness

Some stars are naturally more luminous than others ,so the brightness level from one star to next star is significantly different.

2) The star distance from us

The more distance of an object the dimmer it appears.

Energy emitted = sAT⁴

where s is stefan constant

A is surface area and T is temperature  .  

to learn more about Luminosity click here brainly.com/question/14140223

#SPJ4

7 0
11 months ago
The capacitance of a fully-charged capacitor is 11 F. Determine the capacitor's capacitance when it is half charged.
Tcecarenko [31]

Answer:

The capacitance is 11 F for half and fully charged capacitor.

Explanation:

Capacitance of capacitor is given by the expression

             C=\frac{\epsilon A}{d}

Where ε is the  vacuum permittivity, A is the area of plates and d is the separation between plates.

So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.

Here the capacitance is 11 F for half and fully charged capacitor.

4 0
3 years ago
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