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Sever21 [200]
4 years ago
9

If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

harge and the same amount of negative charge. If you placed these charges 8.00 m apart, how strong would the attractive forces between them be?
Physics
1 answer:
Kryger [21]4 years ago
7 0

Answer:

The attractive force between them is 1.296 \times 10^{18} N

Explanation:

Given:

Charge q = 96000 C

Distance between two charges r = 8 m

According to the coulomb's law,

    F = \frac{kq^{2} }{r^{2} }

Where k = 9 \times 10^{9} = force constant.

   F = \frac{9 \times 10^{9} \times (96000)^{2}  }{8^{2} }

   F = 1.296 \times 10^{18} N

Therefore, the attractive force between them is 1.296 \times 10^{18} N

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Label the layers of the Earth in order from youngest to oldest according to the law of superposition.
zaharov [31]

Answer:

With the increase in depth, age of layer increases.

Explanation:

According to the law of superposition, the layer of the Earth that is present in the deep is considered as the oldest layer while on the other hand, those layer which is present on the top of all layers is considered as youngest layer of earth. When we move from the top layer towards the bottom layer, the age of layer increases or in other words, when we move upward the age of the layer decreases so label the layers of earth on the basis of this phenomenon.

4 0
3 years ago
4. A 70.0 kg boy and a 45.0 kg girl use an elastic rope while engaged in a tug-of-war on an icy,
marta [7]

Answer:

a_1 = 1.446m/s^2

Explanation:

Given

m_1 = 70.0kg -- Mass of the boy

m_2 = 45.0kg -- Mass of the girl

a_2 = 2.25m/s^2 -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl (a_1)

From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:

m_1 * a_1 = m_2 * a_2

Substitute values for m_1, m_2 and a_2

70.0 * a_1 = 45.0 * 2.25

Make a_1 the subject

\frac{70.0 * a_1}{70.0} = \frac{45.0 * 2.25}{70.0}

a_1 = \frac{45.0 * 2.25}{70.0}

a_1 = \frac{101.25}{70.0}

a_1 = 1.446m/s^2

<em>Hence, the acceleration of the boy towards the girl is 1.446m/s^2</em>

7 0
3 years ago
Password is 1234 ....
irina1246 [14]

Answer:

okay

Explanation:

have a nice day :)  stay hydrated.

‧₊˚ ꒰ mei ꒱ ‧₊˚

3 0
3 years ago
Read 2 more answers
A car drives 2 miles north, 4 miles west, 2 miles south and then drives 6 miles east. What is the car's distance?
geniusboy [140]

Answer:

Ok, so the car has traversed a total of 2 + 4 + 2 + 6 = 14 miles.

That's the car's distance! :)

For the displacement, you can draw out a diagram. I'll try my best to make one using a keyboard xDDD

 -    -   -   -    |

 |                  |

 |    -   -    -  start  -   end!

So, this is the journey of the car: Going two up, 4 left, 2 down, and 6 right.

This ends up 2 to the right of the beginning!

That's a 2 mile displacement :)

Can you give me brainliest for my splendid diagram? xD

4 0
3 years ago
Help me please i’ll pay you (paypal)
Arisa [49]

If the award weighs 200 newtons and 200 newtons equals 44.96 pounds of force even though it is of such a force if it hits the ground the energy will either discharge into the air doing nothing but creating a loud sound or it will discharge into the ground altering the ground that it landed on.

Hope this helps :)

8 0
3 years ago
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