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SVETLANKA909090 [29]
2 years ago
6

Which element rarely reacts because it has 8 electrons in its outer shell?

Physics
1 answer:
sashaice [31]2 years ago
3 0

Answer:

only neon has electron i n it's outer shell.so it reacts normally.other given elements do not have 8 electron in its outer shell.

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A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi
Norma-Jean [14]

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

8 0
3 years ago
True Or False;
tino4ka555 [31]
False: because atoms are base on the elements on the periodic table.
3 0
3 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
if the radius of the artery decreases by a factor of 1.5 because of the blockage by what factor has the volume flow rate decreas
Pavel [41]

Answer:

if the radius of the artery decreases by a factor of 1.5 because of the blockage by what factor has the volume flow rate decreased?

8 0
3 years ago
Read 2 more answers
X-ray diffraction is used to study the structure of crystallized proteins, nucleic acids, and other biological macromolecules. I
stepladder [879]

Answer:

PART A: option b. .43nm

PART B: option d. 0.11nm

PART C: option c. The wavelengths of visible light are too long compared to the atomic spacing.

Explanation:

Given data

Wavelength λ = 0.20 nm

Angle θ = 0.8 rad

(a)

wavelength of x-ray to give maximum at the same location

λ₂ = m λ

Here, m = 2 is the interference fringe order.

Substitute the values in the above equation.

λ₂ = 2 × 0.2

    = 0.4 nm

Hence, the wavelength of x-ray to give maximum at the same location is 0.4nm

(b)

The crystal plane separation is equal to d

The value of θ is equal to 0.8 rad.

Convert rad into degree as follows:

0.8 rad = \frac{180^0}{\pi rad} (\frac{0.8rad}{1}) = 144°/π = 45.86°

Solve for d, using equation (1) as follows:

2dsinθ = mλ

d = mλ / 2sinθ

d = (1) 0.17 / 2Sin45.86°

d = 0.17 / 1.9065

d = 0.089 nm

(c)

The visible light can not be used to study the structure of proteins because of the high wavelength of the visible light.

7 0
2 years ago
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