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Vika [28.1K]
3 years ago
12

What are the signs of "Depression"?

Physics
2 answers:
antiseptic1488 [7]3 years ago
7 0
No money people are poor people can not buy anything
Sergio039 [100]3 years ago
6 0
Some signs can be
-not wanting to be around people
-feeling tiered or not wanting to do anything
-not finding fun in thing you used to find enjoyable
-extreme sadness all the time (being depressed)
-suicidal thoughts
-cutting oneself
-questioning if life is really worth living.

from someone who experiences depression daily I feel I have covered a few points. there is much more that goes on but here are some of the basics.
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Answer:

1

Explanation:

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3 years ago
A building that is 105.50 ft. feet tall is casting a shadow that is 37.50 ft. feet. If the building next to it, is casting a sha
Arlecino [84]
115.35 ft

Set the proportion up 37.50/105.50 = 41/x and solve for x
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Determina la velocidad de un avión que recorre 459 km en 67 min
satela [25.4K]

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122397.7

Explanation:

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2 years ago
If you are holding a book and you let go of it, why does it fall? Explain in terms of forces, potential energy, and kinetic ener
inessss [21]

Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

Due to this process the whole energy is conserved.

As the potential energy decreases kinetic energy increases.

3 0
2 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
3 years ago
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