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Leviafan [203]
3 years ago
8

The elements that do not ordinarily form compounds are

Physics
2 answers:
ololo11 [35]3 years ago
8 0

The noble gasses, they are the elements on group 18 of the periodic table and do not usually form compounds. The noble gases are as follows: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

Hope this helped :)

 

pishuonlain [190]3 years ago
7 0

Answer:

Noble gases

Explanation:the have all 8 valence electrons so they do not have a need to bond

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Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
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\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


6 0
3 years ago
A car goes from 5 m/s to 25 m/s in 6 s. What is the acceleration of the car?
damaskus [11]

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3 years ago
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
atroni [7]

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The average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

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The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V

So, the average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

6 0
4 years ago
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Answer:

D

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4 years ago
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