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zzz [600]
4 years ago
13

Which of the following is true about comets? A. The orbits of comets do not take them past the Jovian planets. B. Comets are hel

d together by frozen gases. C. All comets have long tails made of vaporizing gases. D. The Oort cloud of comets is found between Neptune and Pluto.
Physics
2 answers:
Leya [2.2K]4 years ago
8 0
I believe its B. but im not sure
olganol [36]4 years ago
8 0
I chose B and it said it was correct so go with that

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In both ionic and molecular bonds, the resulting compound is stabilized because
frosja888 [35]

Answer: In both ionic and molecular bonds, the resulting compound is stabilized because each atom's outer electronic orbital is full.

Explanation:

Molecular bonds are also called covalent bonds. A covalent bond is formed by sharing of electrons between two or more atoms.

For example, atomic number of hydrogen is 1 and atomic number of nitrogen is 7 (2, 5). In order to attain stability hydrogen atom needs to gain one electron whereas nitrogen needs to gain 3 electrons.

Hence, 3 atoms of hydrogen chemically combine with one atom of nitrogen by sharing electrons and thus it forms the compound NH_{3}.

Ionic bonds are the bonds formed by transfer of electrons from one atom to another.

For example, atomic number of sodium is 11 (2, 8, 1) and atomic number of chlorine is 17 (2, 8, 7). In order to attain stability sodium needs to lose one electron whereas chlorine needs to gain one electron.

Hence, when sodium combines chemically with chloride then sodium will transfer its 1 valence electron to the chlorine atom and thus it forms the compound NaCl.

Therefore, we can conclude that in both ionic and molecular bonds, the resulting compound is stabilized because each atom's outer electronic orbital is full.

7 0
3 years ago
A bicyclist is in a 50-km race. She says she had an average velocity of 35. What is missing in the cyclist's velocity? A. units
aivan3 [116]
Hello There!

It is A. Units only.

Hope This Helps You!
Good Luck :) 

- Hannah ❤
4 0
4 years ago
Read 2 more answers
HELPPP ASAP!!!!!!
fomenos

Answer:

The acceleration of the player is - 4.9 m/s²

Explanation:

The given is:

1. The mass of the player is 55 kg

2. His initial speed is 4.6 m/s

3. The coefficient of the kinetic fraction between the player and the

    ground is 0.50

We need to find the player acceleration

According to Newton's Law

→ ∑ forces in direction of motion = mass × acceleration

There is only the friction force opposite to the motion

→ Friction force = μR

where μ is the coefficient of friction and R is the normal reaction

→ The normal reaction R = mg

where m is the mass and g is the acceleration of gravity

→ m = 55 kg , g = 9.8 m/s²

→ R = 55 × 9.8 = 539 N

→ ∑ F = - μR

→ - μR = m × a

→ μ = 0.5 , R = 539 N , m = 55

→ -(0.5)(539) = 55 × a

→ - 269.5 = 55 a

Divide both sides by 55

→ a = - 4.9 m/s²

The acceleration of the player is - 4.9 m/s²

Learn more:

You can learn more about Newton's law in brainly.com/question/11911194

#LearnwithBrainly

3 0
4 years ago
A ray of light strikes a smooth surface and is reflected. The angle of incidence is 35°. What can be predicted about the angle o
andreev551 [17]
By the law of reflection which says that the angle of reflection is equal to the angle of incidence, which in other words says, sin x = sin y, it can be said, based on the given that the angle of reflection is also equal to the angle of incidence of 35 degrees
5 0
3 years ago
Read 2 more answers
A myopic person has a far point of 1.000m. If she wears her eyeglass 2.0cm in fromt of here eyes, what is the power of her eye g
Iteru [2.4K]

Answer:

51 dioptre

Explanation:

u = Object distance =  2 cm

v = Image distance = 1 m

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{0.02}+\frac{1}{1}\\\Rightarrow \frac{1}{f}=\frac{1.02}{0.02}\\\Rightarrow \frac{1}{f}=51\\\Rightarrow f=\frac{1}{51}

Power of lens

P=\frac{1}{f}\\\Rightarrow P=\frac{1}{\frac{1}{51}}\\\Rightarrow P=51\ D

Power of the eye lens is 51 dioptre.

5 0
3 years ago
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