1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
NNADVOKAT [17]
3 years ago
5

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
Other questions:
  • Which element is most likely to carry electric current easily?
    15·1 answer
  • A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.7 kg and the sign has a mass of ms = 17.
    15·1 answer
  • The magnetic field or force seems to be associated with the lineup of____________ within the magnet.
    11·2 answers
  • At what point in its motion is the KE of a pendulum bob a maximum? 1. The KE does not change. 2. at the lowest point 3. at the h
    7·1 answer
  • How fast is a cat that runs 3 kilometers in 0.5 hours
    14·1 answer
  • A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in
    9·1 answer
  • HELP PLEASE THANKS!! Explain why Gravitational forces are always attractive.
    7·1 answer
  • Here is the picture of an aircraft, with the forces working on it shown. Identify the force that corresponds to the question mar
    10·2 answers
  • If you place a piece of paper containing a small letter "d" on the stage, what will the image look like under the microscope?
    8·2 answers
  • What are the units for specific charge?<br> A. C/kg<br> B. C*kg<br> C. kg/C<br> D. 1/C*kg
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!