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Aleksandr [31]
3 years ago
11

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

een the first and second minima in the diffraction pattern is 4.75 mm. What is the width (in mm) of the slit?
Physics
1 answer:
mario62 [17]3 years ago
3 0

Answer:

0.2 mm

Explanation:

As we know that

Y = \frac{m\lambda * D}{d}

where

m represents  the order of minimum

y represents the  distance on the screen of the minimum from central axis

λ is the wavelength  of the light

D is the distance between  screen-to-slit

d represents the width of the slit

For first minima

y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}

For second minima

y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}

Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm

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Given that the rope is not moving (acceleration is zero), by the second Law of Newton (F=m*a), the net force acting on the rope is zero.

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On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
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2 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
Eduardwww [97]

Answer:

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So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

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3 years ago
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Answer:

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Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\

<u>Magnitude of Vector = 207.73 m</u>

(b)

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Answer: i dont know yet

Explanation:

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