All categories

3 years ago

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron

Physics

2 answers:

Inessa05 [86]3 years ago

8 0

Answer:

providing energy to an atom can allow the electron in its non valence shell to obtain energy and move to a higher energy orbital and act as a valence electron.

Explanation:

liq [111]3 years ago

6 0

Energy-

and it can be added to an atom to cause a non-valence electron in the atom to temporarily become a valence electron

You might be interested in

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago

What is the daily use of the deltoid?

Ghella [55]

Answer:

In our everyday life, the deltoids take on most of the work in rotating our arms, but besides that the deltoid muscle also allows us to carry objects at a safe distance from the body. It is also responsible from stopping a dislocation or injury to the humerus when we carry something heavy.

6 0

3 years ago

Read 2 more answers

What is the acceleration if velocity increases from 10 m/s to 15m/s after travelling a distance of 5 metre

Semenov [28]

Answer:

a=1.25m/s²

Explanation:

GIVEN DATA

vi=10m/s

vf=15m/s

S=5m

TO FIND

a=?

SOLUTION

by using third equation of motion

2as=(vf)²-(vi)²

2a(5m)=(15m/s)²-(10m/s)²

10m×a=225m²/s²-100m²/s²

10m×a=125m²/s²

a= $\frac{125}{10}$

a=12.5m/s²

8 0

3 years ago

What did ancient astronomers think areas of the moon called mares might be?

Bumek [7]

The ancient astronomers think areas of the moon called mares might be Seas.

Option D

<u>Explanation</u>:

The surface area of Earth's moon is dark, large, and is basaltic plains which are formed by ancient volcanic eruptions. They were dubbed as Maria, "ancient astronomers" who misunderstood them as actual seas. They are less reflective than highlands. Due to their iron-rich composition, they tend to appear dark from the naked eye. The Maria cover approximately about 16% of surface mostly on side that is visible from Earth. The few Maria on side that is too far are much smaller and residing mostly in very large craters. The ancient astronomers mistook the surface area as look like actual seas.

4 0

3 years ago

xa student fires a cannonball diagonally with an initial speed of 35.0m/s. neglect drag and the initial height of the cannonball

ICE Princess25 [194]

2.23 seconds is the cannons' total flight time.

What is Acceleration due to gravity ?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/ $s^{2}$ is its SI unit. Both magnitude and direction are present.

According to the information

A diagonal ball is launched at a 45° angle from the horizontal. The ball is being fired in a projectile motion. A projectile is a motion in which a body that has been discharged into space with a velocity U is allowed to fall freely while being affected by gravity.

The cannon ball's total flight time, T, can be represented as;

where T = 2Usin(theta)/g;

U is the ball's starting speed, which is 35 m/s.

The angle the ball makes with the horizontal is theta, which is equal to 45°.

The acceleration caused by gravity is g = 9.81 m/ $s^{2}$

Incorporating the provided data into the existing formula;

T = 35sin45°/9.81

T = 35×0.7071/9.81

T = 24.74/9.81

2.52 seconds later

2.23 seconds is the cannons' total flight time.

To know more about Acceleration due to gravity

brainly.com/question/13860566

#SPJ4

8 0

1 year ago