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3 years ago

What is generally TRUE about diagnosing psychological disorders?

1 answer:

3 0

The statement that says "Psychological disorders can be very difficult to diagnose" is true about diagnosing psychological disorders.

<h2>What are psychological disorders?</h2>

Psychological disorders are those mental, behavioral, emotional and thinking conditions that interfere with the normal performance of the individual in society.

Mental disorders are psychiatric conditions that are expressed in a syndrome, verifiable from different diagnostic criteria.

The steps to obtain a diagnosis include a medical history, physical examination, and possibly laboratory tests and a psychological evaluation.

Therefore, we can conclude that a psychological disorder is an alteration in the mental balance of a person that requires specialized attention adapted to the characteristics of the dysfunction.

Learn more about psychological disorders here: brainly.com/question/6367767

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If the Sun were to collapse into a black hole, the point of no return for an investigator would be approximately 3 km from the c

Marysya12 [62]

Answer:

1.97*10⁴N .He cannot survive.

Explanation:

The<em> </em><em>Tidal force</em> can be defined as the difference in force between center of two bodies like center of earth and other bodies.This force stretches the body too and fro from the center of mass of another body due to the difference in the gravitational force.

The tidal force is responsible for various phenomena like ocean tides,celestial body tearing up.

The tidal force is the effect of massive body gravitationally affecting another body.

The tidal force is the difference between the gravitational force of two bodies and directly proportional to the mass of body affecting another body and inversely proportional to the square of radius of the distance between center and the object.

Here let us consider height of person as 2m and d=1m

ΔF= $\frac{GMm}{(R-d)^{2} }$ - $\frac{GMm}{(R+d)^{2} }$

=GMm[ $\frac{1}{(R-d)^{2} }$ - $\frac{1}{(R+d)^{2} }$ ]

=(6.67*10⁻¹¹)(1.99*10³⁰)(1.0)[ $\frac{1}{(300*10^3-1)^{2} }$ - $\frac{1}{(300*10^3+1)^{2} }$ }

ΔF=1.97*10⁴N.

Thus he cannot survive

4 0

3 years ago

Read the article Watering Livestock with Renewable Energy.

DaniilM [7]

Answer:

D. Wind and solar energy help save money and reduce air pollution.

Explanation:

7 0

3 years ago

Read 2 more answers

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

3 years ago

Question 2 10

ss7ja [257]

B Ik it all just know who they’ll you

8 0

3 years ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

3 years ago