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butalik [34]
3 years ago
13

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo

u set your slit spacing at 1.05 mm and place your screen 8.91 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.97 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?
Physics
1 answer:
Reptile [31]3 years ago
4 0

Answer:

The wavelength is \lambda = 586nm

Explanation:

From the question we are told

       The slit spacing is d =1.05mm = \frac{1.05}{1000} = 1.05*10^{-3}m

       The distance from the screen L = 8.91m

       The distance from central bright fringe y_n = 4.97cm = \frac{4.97}{100} = 0.0497m

       The order of the bright fringe n = 10

Generally the position of a bright  fringe is mathematically represented as

            y_n = n\frac{\lambda L }{d}

            Where  \lambda is the wavelength

             Making \lambda the subject

                        \lambda = \frac{y_n d}{nL}

substituting  value

                      \lambda = \frac{0.0497 * 1.05 *10^{-3}}{10 * 8.91}

                      \lambda = 586nm

     

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Two objects, each with a charge of +0.15 C are separated by a distance of 3 meters.
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6 0
3 years ago
A trailer truck with a 2000 [kg] cab and a 8000 [kg] trailer is traveling on a level road at 90 [km/hr].The brakes on the traile
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Answer:

a)   t = 19.6 s, b) fr = 1.274 10⁴ N

Explanation:

This is a Newton's second law problem

Y Axis

for the cabin

        N₁-W₁ = 0

        N₁ = W₁

for the trailer

        N₂- W₂ = 0

        N₂ = W₂

X axis

for the cabin plus trailer, where friction is only in the cabin

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the friction force equation is

        fr = μ N

we substitute

       μ N₁ = (m₁ + m₂) a

        μ m₁ g = (m₁ + m₂) a

        a = μ g    \frac{m_1}{m_1 + m_2}

         

let's calculate

         a = 0.65 9.8    \frac{2000}{2000+8000}

         a = 1,274 m / s²

a) to find the stopping distance we can use kinematics

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         v = v₀ - a t

when it is stopped its speed is zero

           0 = v₀ - at

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           t = 25 / 1.274

           t = 19.6 s

b) the friction force is

           fr = 0.65 2000 9.8

          fr = 1.274 10⁴ N

This is the braking force and also the forces that couple the cars.

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