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4 years ago

How do you think overpumping groudwater is related to the formation of sinkholes?

1 answer:

vladimir2022 [97]4 years ago

3 0

Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps

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A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled

topjm [15]

average velocity is vector displacement / time

time is "almost exactly one hour"

disp = -10m

v= -10/1x60x60 = -1/360m/s

5 0

3 years ago

Atoms with a low ionization energy give up their outer valence electrons with

Sedbober [7]

Answer:

a. true

Explanation:

i think

that my answer

7 0

3 years ago

What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element

Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0

3 years ago

Read 2 more answers

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

What is the momentum of a 1200 kg car traveling with a speed of 27 m/s (60 mph)?

serious [3.7K]

Answer:

This is your answer

Explanation:

Actually I took this from go ogle

3 0

3 years ago