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3 years ago

7

How has psychology developed from its prescientific roots in early understandings of mind and body to the beginnings of modern

science?

1 answer:

bonufazy [111]3 years ago

3 0

Answer:

By the experiences, ideas of origin, power and modern empiricism.

Explanation:

The route of the psychology has been recorded by the history of India, Middle East, Europe and China. The Confucius and Buddha focused on the ideas origin and the power. The ancient Hebrews, Aristotle and Pluto tells the idea whether the body with the mind is connected or it is distinct and this connection or distinct idea is the result of experiences.

Locke offers his theories about the mind that mind as the blank state, on which the experience writes. The modern empiricism idea has been developed by Bacon and Locke

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Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?

Lera25 [3.4K]

Answer:

The value is $w = 0.1167 \ rev/second$

Explanation:

From the question we are told that

The rate at which the plate rotates is $w =7.0 \ rev/min$

Generally the revolution per second is mathematically represented as

$w = \frac{7.0}{60}$

=> $w = 0.1167 \ rev/second$

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2 years ago

Why is it important to be able to trace the pole connection on a meter back to the same type of pole at the electrical source?

lozanna [386]

Answer:

Explanation:

A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.

In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.

When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.

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3 years ago

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

3 years ago

un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración

sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

Vf: Velocidad final
Vo: Velocidad inicial
a: Aceleración
d: Distancia recorrida

En este caso:

Vf: 0 m/s, porque el avión se detiene
Vo: 50 m/s
a: ?
d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

$a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}$

a= - 10.42 m/s²

<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>

5 0

3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

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