Answer:
Explanation:
Width of central diffraction peak is given by the following expression
Width of central diffraction peak= 2 λ D/ d₁
where d₁ is width of slit and D is screen distance and λ is wave length.
Width of other fringes become half , that is each of secondary diffraction fringe is equal to
λ D/ d₁
Width of central interference peak is given by the following expression
Width of each of bright fringe = λ D/ d₂
where d₂ is width of slit and D is screen distance and λ is wave length.
Now given that the central diffraction peak contains 13 interference fringes
so ( 2 λ D/ d₁) / λ D/ d₂ = 13
then ( λ D/ d₁) / λ D/ d₂ = 13 / 2
= 6.5
no of fringes contained within each secondary diffraction peak = 6.5
Scobie will take 10 days to drive around Earth's equator.
To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:
<em>Where:</em>
r: is the Earth's radius = 6371 km
Then, the distance is:
Now, if we divide the above distance by the speed of the car we can find the time:
Therefore, Scobie will take 10 days to drive around Earth's equator.
To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults
I hope it helps you!
Answer:
1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.
2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.
3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
