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3 years ago

What is the total number of orbitals containing only one electron in an atom of nitrogen in the ground state?

2 answers:

6 0

C is the answer. nitrogen is in subshell p and this subshell can hold 6 electrons therefore it has 3 orbitals. since nitrogen has three electrons in this subshell, only 1 electron is in each shell. hope that helps.

Sliva [168]3 years ago

4 0

Answer: 3

Explanation:

Nitrogen is in subshell p which can only hold 6 electrons, therefore, it has three orbitals. Since Nitrogen has three electrons in this subshell, only one electron is in each orbital.

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Refraction occurs when a wave changes its A) color. B) frequency. C) intensity. D) speed.

Gwar [14]

The answer would be d. speed.

3 0

3 years ago

Read 2 more answers

The drag force on a falling coffee filter can be modeled as a linear function with respect to velocity, F⃗ D =−bv, where b is a

nordsb [41]

Newton's second law and graphical analysis allow us to find the correct answer for the measure of the constant b is:

D) Stack several filters to change the mass measure the acceleration of the system.

Newton's second law establishes a relationship between the net force, the mass and the acceleration of the bodies.

∑ F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.

They indicate that the drag force on the filters is

F = - b v

Where b is a positive constant that depends on the shape and area of the filter and v is the speed of the filter.

Let's write Newton's second law.

W - fr = ma

W -bv = ma

In the experiment the students indicate that they can measure the position, velocity and acceleration of the body.

Based on the above, if we place several filter weights and measure their speed and acceleration in each case, we can make a graph of velocity versus acceleration, we can take the value of the constant b from the slope.

Let's analyze the different answers:

A) False. The constant b depends on the shape of the filter therefore it must be kept constant.

B) False. The constant depends on the area, so it must be kept constant.

C) May be. In this case, since we have the terminal velocity, the acceleration is zero, Newton's second law remains.

B v - W = 0

b = $\frac{W}{v}$

The problem with the method is the difficulty of measuring the compression terminal velocity.

D) True. According to the discussion of the velocity versus acceleration. graph, the constant b is equal to the slope of the graph.

E) May be. The problem with this method is finding a reliable value for the terminal velocity.

In conclusion, using Newton's second law and graphical analysis we can find the correct answer for the measure of the constant b is:

D) Stack several filters to change the mass measure the acceleration of the system.

Learn more here: brainly.com/question/2441565

6 0

2 years ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

2 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago