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shepuryov [24]
3 years ago
8

Suppose you had the same laser and diffraction grating from the previous question but now you had a flat detection screen. You w

ant to use this set up to measure the displacement due to small vibrations of the detection screen. (The vibrations cause the screen to move closer and farther from the grating.) A. Describe how you would do measurements for this experiment to determine the displacement of the detection screen.
Physics
1 answer:
kolezko [41]3 years ago
4 0

Answer:

measuring the zero intensity point, we can deduce the movement of the screen.

The distance from the center of the pattern to the first zero is proportional to the distance to the screen,

Explanation:

The expression for the diffraction phenomenon is

           a sin θ = m λ

for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles

           tan θ = y / L

     

in these experiments the angles are small

          tan θ = sin θ / cos θ = sin θ

          sunt θ = y / L

we substitute

          a \frac{y}{L}= m  λ

           y = m L λ / a

therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.

 

The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.

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Answer:

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4 0
3 years ago
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An 1800 kg helicopter rises with an upward acceleration of 2.0 m/s?. What lifting force is supplied by its rotating blades?
Viktor [21]

Answer:

Lifting force, F = 21240 N

Explanation:

It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force  supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, F=1800\ kg(9.8\ m/s^2+2\ m/s^2)

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

4 0
3 years ago
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/
Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0
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allsm [11]

Answer:

D) Grounding

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5 0
3 years ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

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for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
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