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2 years ago

Client-centered therapy, Gestalt therapy, and psychoanalysis are examples of ________ therapies.

Physics

1 answer:

expeople1 [14]2 years ago

4 0

Client-centered therapy, Gestalt therapy and Psychoanalysis are all examples of psychotherapy.

<h3>What is therapy?</h3>

Therapy is the process of working as a counselor to develop positive thinking and treat specific mental health conditions like depression or anxiety, trauma. Psycho word is all meant for emotions.

Psychoanalysis, client-centered therapy, Gestalt therapy, rational-emotive therapy, and cognitive-behavior therapy are all examples of psychotherapy.

Thus, Client-centered therapy, Gestalt therapy and Psychoanalysis are all examples of psychotherapy.

Learn more about therapy.

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No waves because Q19 waves would going at the surface at regions

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3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
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3 years ago

How do you identify the number of valence electrons in an element using periodic table

frosja888 [35]

You may look at what group they are in
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5 0

3 years ago

Read 2 more answers

A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horiz

Nataliya [291]

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

$F_s = \mu_s mg$

where

$\mu_s$ is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

$F_s=73 N$

Using this information into the previous equation, we can find the coefficient of static friction:

$\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355$

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

$F_k = \mu_k mg$ (1)

where

$\mu_k$ is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

$F-F_k = ma = 0$

And by substituting (1), we can find the value of the coefficient of kinetic friction:

$F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267$

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4 years ago

Which of the following is an example of a safe laboratory produced?

Step2247 [10]

I think you meant to say "procedure" instead of "produced".

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3 years ago