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ollegr [7]
3 years ago
13

The diagram shows two different types of fossils from the

Physics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

I think the answer is A. X: Mold Y: Cast

Explanation:

Hope that helps!!!

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Rainbow [258]

Answer:

They are all s

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Which Lewis dot diagram shows an atom that needs 2 more electrons in its outermost shell?
sergiy2304 [10]

Answer: Option (c) is the correct answer.

Explanation:

Atomic number of hydrogen is 1 and one dot in its lewis structure displays only one electron. So, in order to complete its octet it needs one more electron.

Atomic number of carbon is 6 and its electronic distribution is 2, 4. Carbon needs 4 more electrons in order to complete its octet.

Atomic number of oxygen is 8 and its electronic distribution is 2, 6. In order to complete octet, oxygen needs 2 more electrons.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. In order to complete octet, fluorine needs 1 more electron.

Hence, we can conclude that out of the given options, lewis dot structure of oxygen atom needs 2 more electrons in its outermost shell.

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4 years ago
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What is a Harmful mutation
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3 years ago
A baseball is located at the surface of the earth. which statements about it are correct? (there may be more than one correct ch
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Answer:The gravitational force on the ball due to earth is exactly the same as the gravitational force on earth due to the ball

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If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
Dennis_Churaev [7]

Answer:

W' = 1.66 x 10¹⁴ N

Explanation:

First, we will calculate the mass:

W = mg

where,

W = weight on earth = 690 N

m = mass = ?

g = acceleration due to gravity on earth = 9.8 m/s²

Therefore,

m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg

Now, we will calculate the value of g on the neutron star:

g' = \frac{GM}{R^2}

where,

g' = acceleration due to gravity on the surface of the neutron star = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of the Neutron Star = 1.99 x 10³⁰ kg

R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m

Therefore,

g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2

Therefore, the weight on the surface of the neutron star will be:

W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)

<u>W' = 1.66 x 10¹⁴ N</u>

7 0
3 years ago
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