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3 years ago

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In the early part of the 1900s, psychologists who broke down thought processes into their basic elements and analyzed them were

called structuralists. What term might we use to describe
psychologists with similar interests today?
Behavioral
Cognitive
Gestalt
Humanistic
Social

1 answer:

Svetlanka [38]3 years ago

8 0

Answer: Cognitive

Explanation:

Cognition can be defined as the process of obtaining the knowledge and learning through the experience, senses and thought. Many psychologists can have similar experiences of learning the psychological framework of the society or at the individual level. Thus this can be said that the cognitive psychologist may have similar interests.

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Why may some young adults go through a quarter-life crisis?

slavikrds [6]

Stress. Being alone. having nothing

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3 years ago

A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the

Darya [45]

The acceleration of the car is $1.2 m/s^2$ upward

Explanation:

We can solve the question by using Newton's Second Law of motion, which states that:

$F=ma$

where

F is the net force exerted on an object

m is the mass of the object

a is its acceleration

In this problem, we know that:

F = 900 N is the net force exerted on the car

m = 750 kg is the mass of the car

Solving the equation for a, we find the acceleration of the car:

$a=\frac{F}{m}=\frac{900}{750}=1.2 m/s^2$

and it is in the same direction as the force (upward).

Learn more about Newton's second law of motion here:

brainly.com/question/3820012

#LearnwithBrainly

8 0

3 years ago

A 6.00kg box is subjected to a force F=18.0N-(0.530N/m)x. Ignoring friction and using Work, find the speed of the box after it h

Nadusha1986 [10]

Answer:

Approximately $8.17\; \rm m \cdot s^{-1}$ assuming that the effect of gravity on the box can be ignored.

Explanation:

If the force $F$ is constant, then the work would be found with $W = F \cdot \Delta x$ . However, this equation won't work for this question since the

$\displaystyle W = \int\limits_{x_0}^{x_1} F\, d x$ ,

For this particular question, $x_0 = 0\; \rm m$ and $x_1 = 14.0\; \rm m$ . Apply this equation:

$\begin{aligned}W &= \int\limits_{x_0}^{x_1} F\, d x \\ &= \int\limits_{0\; \rm m}^{14.0\; \rm m} \left[{18.0\; \rm N} - {\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x \right]\, d x \\ &= \left[{(18.0\; \rm N)}\cdot x - \frac{1}{2}\;{\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x^2\right]_{x = 0\; \rm m}^{x = 14.0\; \rm m} \approx 200.06\; \rm N \cdot m\end{aligned}$ .

(Side note: keep in mind that $1\; \rm J = 1\; \rm N\cdot m$ .)

Since friction is ignored, all these work should have been converted to the mechanical energy of this object.

Assume that the effect of gravity on this box can also be ignored. That way, there won't be a change in the gravitational potential energy of this object. Hence, all these extra mechanical energy would be in the form of the kinetic energy of this box.

That is:

$\begin{aligned}& \text{Kinetic energy of this object} \\ =& \text{Initial Kinetic Energy} + \text{Change in Kinetic Energy} \\ =& \text{Initial Kinetic Energy} + \text{Change in Mechanical Energy} \\ =& \text{Initial Kinetic Energy} + \text{External Work} \\=& 0\; \rm N \cdot m + 200.06\; \rm N \cdot m \\ =& 200.06\; \rm N \cdot m \end{aligned}$ .

Keep in mind that the kinetic energy of an object of mass $m$ and speed $v$ is:

$\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Therefore:

$\begin{aligned}v &= \sqrt{\frac{2\, (\text{Kinetic energy})}{m}} \\ &= \sqrt{\frac{2\times 200.06\; \rm N \cdot m}{6.00\; \rm kg}} \approx 8.17\; \rm m \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

KiRa [710]

Answer:

(a) Determine the astronaut's orbital speed

The velocity of the astronaut is 1436m/s.

(b) Determine the period of the orbit.

The period of the orbit is 10413 seconds.

Explanation:

<em>(a) Determine the astronaut's orbital speed</em>

The orbital speed of astronaut can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Then, replacing Newton's second law in equation 3 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a is the centripetal acceleration since the astronaut describes a circular motion around the Moon:

$a = \frac{v^{2}}{r}$ (3)

Replacing equation 3 in equation 2 it is gotten:

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

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

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

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (4)

Where v is the orbital speed, G is the gravitational constant, M is the mass of the Moon, and r is the orbital radius.

Notice that the orbital radius will be given by the sum of the radius of the Moon and the height of the astronaut above the surface.

But it is necessary to express the height of the astronaut above the surface in units of meters before it can be used.

$r = 680km \cdot \frac{1000m}{1km}$ ⇒ $680000m$

$r = 1.70x10^{6}m+680000m$

$r = 2380000m$

However it is neccesary to find the mass of Moon in order to use equation 4.

Then Newton's second law ( $F = ma$ ) will be replaced in equation (1):

$ma = G\frac{Mm}{r^{2}}$

Then, M will be isolated

$M = \frac{r^{2}a}{G}$ (5)

Where r is the orbital radius of the astronaut and a is the acceleration due to gravity.

$M = \frac{(2380000m)^{2}(0.867 m/s^{2})}{6.67x10^{-11}N.m^{2}/kg^{2}}$ (5)

$M = 7.36x10^{22}Kg$

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(7.36x10^{22}Kg)}{2380000m}}$

$v = 1436m/s$

Hence, the velocity of the astronaut is 1436m/s.

<em>(b) Determine the period of the orbit. </em>

The period of the orbit can be determined by the next equation:

$v = \frac{2\pi r}{T}$ (6)

Where v is the orbital velocity, r is the orbital radius and T is the period of the orbit.

Then, T can be isolated from equation 6.

$T = \frac{2\pi r}{v}$ (7)

$T = \frac{2\pi (2380000m)}{1436m/s}$

$T = 10413s$

Hence, the period of the orbit is 10413 seconds.

6 0

3 years ago

27 points for answering this question

photoshop1234 [79]

It is the second one, it is longitude and moving up and down so its B.

5 0

3 years ago