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Nikitich [7]
2 years ago
15

A. Use the graph and the element made in question 2 to determine the mass of the star.

Physics
1 answer:
Minchanka [31]2 years ago
4 0
Black balls with blue waffles
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An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s
ad-work [718]

Answer:

a

 \theta  =  0.0022 rad

b

 I  =  0.000304 I_o

Explanation:

From the question we are told that  

   The  wavelength of the light is \lambda  = 550 \ nm  =  550 *10^{-9} \ m

    The  distance of the slit separation is  d = 0.500 \ mm = 5.0 *10^{-4} \ m

 

Generally the condition for two slit interference  is  

     dsin \theta =  m \lambda

Where m is the order which is given from the question as  m = 2

=>    \theta  =  sin ^{-1} [\frac{m \lambda}{d} ]

 substituting values  

      \theta  =  0.0022 rad

Now on the second question  

   The distance of separation of the slit is  

       d =  0.300 \ mm  =  3.0 *10^{-4} \ m

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      I  =  I_o  [\frac{sin \beta}{\beta} ]^2

Where  \beta is mathematically evaluated as

       \beta  =  \frac{\pi *  d  *  sin(\theta )}{\lambda }

  substituting values

     \beta  =  \frac{3.142  *  3*10^{-4}  *  sin(0.0022 )}{550 *10^{-9} }

    \beta  = 0.06581

So the intensity is  

    I  =  I_o  [\frac{sin (0.06581)}{0.06581} ]^2

   I  =  0.000304 I_o

3 0
3 years ago
The sum of two component vectors is referred to as the
blondinia [14]
Resultant is the correct answer!
8 0
2 years ago
The highest element in the hierarchical breakdown of the wbs is
Makovka662 [10]
Work package. Hope this helps!
5 0
3 years ago
The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of
jasenka [17]

1) Frequency: 3.29\cdot 10^{15}Hz

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

E=2.18 aJ=2.18\cdot 10^{-18} J

The energy of a photon is given by

E=hf

where h=6.63\cdot 10^{-34}Js is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz


2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

\lambda=\frac{c}{f}

where c=3\cdot 10^8 m/s is the speed of light and f is the frequency. Substituting the frequency, we find

\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm

5 0
2 years ago
If an object undergoes a change in momentum of 10 kg m/s in 3 s ,then the force acting on it is
Paha777 [63]

Answer:

Force = 3.333 Newton

Explanation:

Given the following data;

Change in momentum = 10 Kgm/s

Time = 3 seconds

To find the force acting on it;

In Physics, the change in momentum of a physical object is equal to the impulse experienced by the physical object.

Mathematically, it is given by the formula;

Force * time = mass * change in velocity

Impulse = force * time

Substituting into the formula, we have;

10 = force * 3

Force = 10/3

Force = 3.333 Newton

8 0
2 years ago
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