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poizon [28]
3 years ago
8

Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

their diffraction patterns to be resolved by Rayleigh's criterion
Physics
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

The distance is  D  =  0.000712 \ m

Explanation:

From the question we are told that

    The wavelength of  the  light source is  \lambda  =  700 \ nm = 700 *10^{-9} \  m

     The distance from a pin hole is  x  =  9\ m

       The  diameter of the pin  hole is  d =  1.2 \ mm  =  0.0012 \ m

     

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              D  =  \frac{1.22 \lambda }{d }

substituting values

             D  =  \frac{1.22 * 700 *10^{-9} }{ 0.0012 }

             D  =  0.000712 \ m

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elements

Explanation:

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What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 475 turns, its
VMariaS [17]

Answer:

Output voltage is 1.92 volts.

Explanation:

Given that,

Number of turns in primary coil, N_p=475

Number of turns in secondary coil, N_s=8

Input voltage, V_i=114\ V

We need to find the voltage output of a transformer used for rechargeable flashlight batteries. For a transformer, the number of turns and the voltage ratio is given by :

\dfrac{N_p}{N_s}=\dfrac{V_i}{V_o}\\\\\\V_o=114\cdot\dfrac{8}{475}\\\\V_o=1.92\ V

So, the output voltage is 1.92 volts.

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What can result if an earthquake causes a sudden vertical change in the sea floor?
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If the radius of the equipotential surface of the point charge is 14.3 m at a potential of 2.20 kV, what will be the magnitude o
PSYCHO15rus [73]

Answer:

The magnitude of the point charge is 3.496 x 10⁻⁶ C

Explanation:

Given;

radius of the surface, r = 14.3 m

magnitude of the potential, V = 2.2 kV = 2,200 V

The magnitude of the point charge is calculated as follows;

V = (\frac{1}{4\pi \epsilon _0} )(\frac{Q}{r} )\\\\V = \frac{KQ}{r} \\\\Q = \frac{Vr}{K} \\\\Q = \frac{2,200 \times 14.3}{9\times 10^9} \\\\Q = 3.496 \times 10^{-6} \ C\\\\Q = 3.496 \ \mu C

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6 0
2 years ago
(1
mart [117]

Answer:

The boat should point in angle to the perpendicular to the shore is, Ф = 12° 22'

Explanation:

Given data,

The velocity of the boat, V = 8.2 m/s

The velocity of the flow of the river, V' = 1.8 m/s

Let Ф formed by the boat perpendicular to the shore,

The two velocity components along the perpendicular axes are V and V'

Therefore,

                                  tan Ф = V' / V

                                             = 1.8 / 8.2

                                             = 0.2195

Therefore,

                                         Ф  = tan⁻¹ (0.2195)

                                              = 12° 22'

Hence, the boat should point in angle to the perpendicular to the shore is, Ф = 12° 22'

7 0
3 years ago
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