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Pavlova-9 [17]
9 months ago
8

what seems to happen to the one electron in the outer layer of sodium when it combines with chlorine to form sodium chloride?

Physics
1 answer:
Hunter-Best [27]9 months ago
3 0
It jumps to an excited state
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What two energies does a bulldozer use​
sesenic [268]
Mechanical and electrical
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2 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
2 years ago
A bolt falls off an airplane high above the ground. How far does the bolt have to fall before its speed reaches 100m/s (about 20
omeli [17]
We can use the equation vf (the final velocity) =vi (the initial velocity) +at (aceleration times time)

We know the final velocity 100m/s, the initial velocity 0, and the acceleration (gravity) 9.8m/s^2. So, 100=0+9.8t. t=100/9.8
3 0
2 years ago
what pelagic layer is located in the uppermost limits of the permanent darkness and reflects sound pulses?
Nikitich [7]

Answer:

​deep scattering layer

Explanation:

I hope this is right, I kind of learned this like a year ago

8 0
1 year ago
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