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maksim [4K]
3 years ago
8

Both atypical and traditional antipsychotics _____ levels of _____.

Physics
1 answer:
LiRa [457]3 years ago
5 0

Answer:

Both atypical and traditional antipsychotics <u>blocks</u> levels of<u> </u><u>serotonin</u><u>.</u>

Explanation:

These drugs are capable of reducing or turning off positive psychotic symptoms such as hallucinations, delusions, and language and behavior disorder.

In addition to also acting as mood stabilizers, they have an impact on mania, depression, and produce low anxiety.

These anti-psychotics work on the 5-HT2A receptor, it belongs to the family of serotonin receptors and it is a G protein-coupled receptor. 5-HT is short for 5-hydroxy-tryptamine, the chemical name for serotonin.

This receptor became notorious for its importance as a target for drugs like LSD. Later it became important again due to the action of many antipsychotic drugs, especially atypical ones.

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20 POINTS!
CaHeK987 [17]
Potential energy is the answer
6 0
4 years ago
PLZ EXPLAIN IM SO CONFUSED AND THIS IS DUE TONIGHT. I WILL GIVE 50 POINTS!
bezimeni [28]

When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same. 

I hope this helps. :)

Brainliest?

8 0
3 years ago
John throws a baseball from a standing position. Which of the three possible paths will the ball follow?
Marysya12 [62]

Answer: C

Explanation:

Bro this is common sense. Throw something across the room and watch it. It isnt going to be repulsed and fly up. It also isnt going to continue on forever. It will clearly fall and hit the ground

3 0
3 years ago
Read 2 more answers
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
Oxana [17]

Answer:

A)  I_{total} = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

     I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

      I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

    I = I_{cm} + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

       I_{body} = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

    I_{total} = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0
4 years ago
An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele
Citrus2011 [14]

Answer:

The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

\tau=I\times\alpha

Here, I = mk²

\tau=mk^2\times\alpha

Put the value into the formula

\tau=3\times(0.2)^2\times0.5

\tau=0.06\ N-m

\tau=6.0\times10^{-2}\ N-m

Hence, The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

7 0
4 years ago
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