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2 years ago

What are the signs of "Depression"?

Physics

2 answers:

antiseptic1488 [7]2 years ago

7 0

No money people are poor people can not buy anything

Sergio039 [100]2 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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A train travels 77 kilometers in 4 hours, and then 76 kilometers in 2 hours. What is its average speed?

Masja [62]

( (77/4) + 76/2 )/2 = 28.625 km/h is what i got

4 0

3 years ago

If the original pressure is 5 atm and original volume is 100 L, and the new volume is 20 L, what is the new pressure?

Nana76 [90]

P1V1=P2V2
5*100=P2*20
500=20P
P=25

7 0

3 years ago

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

5 0

3 years ago

An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the

OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

$m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2$

Through the aforementioned formula we will have to

$v_f^2-v_i^2 = 2ax$

The particulate part of the rest, so the final speed would be

$v_f^2 = 2gx$

$v_f=\sqrt{2(9.8)(3.1)}$

$v_f = 7.79m/s$

Now from Newton's second law we know that

$F = ma$

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

$F = m\frac{dv}{dt}$

Replacing we have that,

$F = (89)\frac{7.79}{0.5}$

$F = 1386.62N$

Therefore the force that the water exert on the man is 1386.62

3 0

3 years ago

A point charge Q of mass M is located initially at the center of the ring. When it is displaced slightly, the point charge accel

dolphi86 [110]

Answer:

n musical notation, stems are the, "thin, vertical lines that are directly connected to the [note] head." Stems may point up or down. ... There is an exception to this rule: if a chord contains a second, the stem runs between the two notes with the higher being placed on the right of the stem and the lower on the left.

Explanation:

6 0

3 years ago