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enyata [817]
3 years ago
9

Which of the following is a mood disorder?

Physics
2 answers:
Sholpan [36]3 years ago
8 0

Answer:

bipolar disorder

Explanation:

bipolar disorder, is previously known as manic depression, is a mental disorder characterized by periods of depression and periods of abnormally elevated mood that last from days to weeks.

gulaghasi [49]3 years ago
5 0
Bipolar disorder is the answer
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Why caring a body and moving with it is not a work done<br><br> Why :​
Alecsey [184]

Answer:

Work done on an object is equal to

FDcos(angle).

So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.

However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.

But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?

Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.

Explanation:

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This object was observed in 1997.<br><br> The object in the picture is ____ a/an .
melisa1 [442]

Answer: Hale-Bopp was an unusually bright comet that flew by Earth, reaching its closest approach to the planet in 1997.

Explanation:

Hale-Bopp is the answer

4 0
2 years ago
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A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along
Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

F(x)=(3\ N/m^2)x^2+(6\ N/m)x

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

W=\int\limits {F{\cdot} dx}

W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

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3 years ago
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What is the correct formula for of the weight w of an object? assume that its mass is m, the magnitude of its acceleration is a,
Vsevolod [243]
Weight = Mass × Acceleration
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