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Svetradugi [14.3K]
4 years ago
6

A space shuttle takes off from Florida and circles Earth several times, finally landing in California. While the shuttle is in f

light, a photographer flies from Florida to California to take pictures of the astronauts when they step off the shuttle. Who undergoes the greater displacement, the photographer or the astronauts?
Physics
1 answer:
stealth61 [152]4 years ago
6 0

Answer:

The displacement of the photographer and the astronauts are equal

Explanation:

In this exercise, it is important to differentiate between distance and displacement.

The distance moved by an object describes the total length of change in position irrespective of the objects starting point or the point where the transit stops. Distance is a scalar quantity as it does not specify a direction of which a transit takes place as such a piston moving in oscillatory motion can be said to cover a very long distance considering its number of completed cycles

Displacement unlike distance is a vector quantity which accounts for an objects starting and ending points regardless of the route through which the object completed the distance, as such it is the straight line linking the object starting and stopping points, of which the length of the line is the magnitude of displacement and the direction from the starting point to the end point

As such in the above question the displacement of the photographer is his transit from Florida to California while the displacement of the astronauts is also from Florida and stopping at California both of which are equal

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Answer:

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kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
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So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

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