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UkoKoshka [18]
3 years ago
11

All of the following could describe an individual with antisocial personality disorder except __________. A. impulsive B. manipu

lative C. aggressive D. responsible Please select the best answer from the choices provided A B C D
Physics
1 answer:
eimsori [14]3 years ago
7 0

D. responsible

Explanation:

Individuals with antisocial personality disorder are not responsible. They are often aggressive, impulsive and quite manipulative.

Antisocial personality disorder is a mental disorder in which a person violates the rights of others without feeling any remorse for doing so.

  • People with such disorders have a high tendency of breaking rules and laws in the society.
  • They are often manipulative and tends to make the society favor them.
  • They carryout activities without a forethought i.e impulsive.
  • They live a chaotic and irresponsible life feeling not duty bound to the society.
  • People with this disorder can be aggressive too.

Often times, this problem can be remedied by therapy

learn more:

Psychological disorder brainly.com/question/4211983

#learnwithBrainly

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3 years ago
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Which nucleus completes the following equation?
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3 years ago
A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
zzz [600]

Answer:

0.0257259766982 m

Explanation:

P_2 = Atmospheric pressure = 101325 Pa

d_1 = Initial diameter = 1.5 cm

d_2 = Final diameter

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h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

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8 0
3 years ago
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saw5 [17]

Answer:

<em>Height = 5.65 km</em>

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