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maw [93]
2 years ago
7

Which statement correctly describes the electrons in a water molecule?. . A.Which statement correctly describes the electrons in

a water molecule?. B.Electrons are pulled closer to the oxygen atom.. C.Electrons are located inside the nucleus of the hydrogen atoms.. D.Electrons are an equal distance between hydrogen and oxygen.
Physics
2 answers:
Zepler [3.9K]2 years ago
7 0

Answer: Option (B) is the correct answer.

Explanation:

Molecular formula of water is H_{2}O. In a water molecule, hydrogen and oxygen ions are held together by strong hydrogen bonding.

Out of hydrogen and oxygen, oxygen is more electronegative. Thus, it will attract the electrons of hydrogen atom more towards itself. As a result, a partial positive charge will develop on hydrogen atom and a partial negative charge will develop on oxygen atom.

Thus, we can conclude that the statement electrons are pulled closer to the oxygen atom correctly describes the electrons in a water molecule.

exis [7]2 years ago
4 0

The statement “Electrons are pulled closer to the oxygen atom” correctly describes the electrons in a water molecule. The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Whipple is confused about the connection between the velocity and acceleration of the tennis ball. he decides to compare the vel
tamaranim1 [39]

The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.

7 0
2 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

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

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord
kiruha [24]

Answer:

Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)

Explanation:

4 0
2 years ago
If a scale on Earth reads 650 N, what is your mass?
OlgaM077 [116]
If the scale reads 650N, then the mass of whoever it is standing on the scale is

         (weight) / (gravity)  =  (650N) / (9.8 m/s²)  =  66.3 kilograms  .

It's not MY mass, even if I'm the one standing on the scale. 
If I stand on a scale and it reads 650 N, the scale is broken.
4 0
2 years ago
A 120-kg refrigerator that is 2.0 m tall and 85 cm wide has its center of mass at its geometrical center. You are attempting to
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to Force of Friction and Torque given by the kinematic equations of motion.

The frictional force by definition is given by

F= \mu mg

Our values are here,

\mu=0.3

m=120kg

g=9.8m/s^2

Replacing,

F=0.30*120*9.8 = 352.8N

Consider the center of mass of the body half its distance from the floor, that is d = 0.85 / 2 = 0.425m. The torque about the lower farther corner of the refrigerator should be zero to get the maximum distance, then

F*x = mg*d

Re-arrange for x,

x= \frac{mg*d}{F}

x= \frac{mg*d}{\mu mg}

x= \frac{d}{\mu}

x= \frac{0.425}{0.3}

x = 1.42m

Then we can conclude that 1.42m is the distance traveled before turning.

6 0
2 years ago
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