__________ and __________ both heavily relied on dream analysis in their treatment of patients. A. Alfred Adler . . . Albert Ellis B. Alfred Adler . . . Carl Jung C. Sigmund Freud . . . Carl Jung D. Sigmund Freud . . . Alfred Adler

Answers

Answer 1

Answer:

C. Sigmund Freud . . . Carl Jung

Explanation:

edge 2021

Answer 2

C.Sigmund Freud and Carl Jung both heavily relied on dream analysis in their treatment of patients.

What is Freud most famous for?

Freud is well-known for inventing and developing the approach of psychoanalysis; for articulating the psychoanalytic idea of motivation, intellectual infection, and the structure of the unconscious; and for influencing medical and popular conceptions of human nature by using positing both everyday and strange thought.

Sigmund Freud was an Austrian neurologist who's perhaps maximum known as the founding father of psychoanalysis. Freud advanced a fixed of therapeutic strategies centered on communication therapy that worried the use of techniques that include transference, loose affiliation, and dream interpretation.

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Related Questions

a sharp image is formed when light reflects from a

Answers

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

What is regular reflection?

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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According to Newton's first law, an object at rest will _____.

never move
stay at rest forever
start moving
stay at rest unless moved by force

Answers

Stay at rest unless moved my force! :)

NEED HELP ASAP, ILL GIVE YOU BRAINLIEST IF CORRECT (30POINTS)


Drag each label to the correct location on the image. Each label can be used more than once.

Identify the parts of the barred spiral galaxy.

SPIRAL ARM, NUCLEUS, BAR


NOTE I JUST FILLED IN THE SPOTS FOR YOU TO SEE, THEY ARE NOT CORRECT

Answers

Answer:

the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space

Explanation:

Answer:

look pkch

Explanation:

A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3​

Answers

Answer:

75.4

Explanation:

r= 2

h= 6

v= 22/7 *r*r*h

v= 75.42

During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent

Answers

Answer:

they are right it is a new moon

Explanation:

took the test

३.रात में घूमने वाला write one word substitute ​

Answers

Explanation:

रात में घूमने वाला arthaarat निशाचर

3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.
A. Which bullet has more kinetic energy?
B. When you double the mass, what happens to the kinetic energy?

Answers

Answer:

A. The bullet with 0.006kg has more energy

B. When the mass is doubled the kinetic energy increases

Explanation:

Kinetic energy increases when mass increases

kinetic energy increases when velocity increases

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm

Answers

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System​

Answers

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.

Answers

Answer:

Δθ = 15747.37 rad.

Explanation:

The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

       [tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]

Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

       [tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]

Solving for Δθ in (2):

       [tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]

The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

       [tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]

Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

       [tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]

Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      [tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]

The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.

it takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. what is the radius of a track?

Answers

Answer:

534.8 meters

Explanation:

Use T=(2*pi*r)/v

560=(2*pi*r)/6

3360=2*pi*r

1680=pi*r

534.8 meters=radius

It takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. The radius of a track is 534.7 m.

What is Distance?The distance covered by a body is equal to the sum of total path covered. It is equal to the total path traveled by an object during its entire journey.This quantity is always positive. It can't be 0 or a negative number.It is defined as a scalar quantity.

Mathematically, it can be calculated as follows :

distance = speed × time

The formula relating distance (d), speed (s), and time (t) is

d = st

First, Calculating the distance,

d = 560 s × 6 m·s⁻¹

  = 3360 m

When, Calculating the track radius,

The distance travelled is the circumference of a circle,

C = 2пr

r = 3360/2п

 = 534.7 m

The radius of the track is 534.7 m.

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On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?

Answers

Answer: Letter B! Is your answer

A reaction occurs when a compound breaks down. This reaction has one reactant and two or more products. Energy, as from a battery, is usually needed to break the compound apart.

Answers

Answer:

decomposition

Explanation:

A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8

Answers

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

a place where two bones come together is known as an

Answers

Answer:

a place where two bones come together is known as a join

Answer:

Hey mate.....

Explanation:

This is ur answer....

Joints

Joints – A place in the body where bones come together. Non-Moveable Joints (sometimes called fixed or fibrous) – A place in the body where two or more bones come together but do not move.

Hope it helps!

mark me brainliest plz....

Follow me! :)

a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.​

Answers

?.............................

. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact

Answers

Answer:

37.5 m/s

Explanation:

Using,

Formula

v = ωr....................... Equation 1

Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.

From the question,

Given: ω = 30 rad/s, r = 1.25 m

Substitute these values into equation 1

v = 30(1.25)

v = 37.5 m/s.

Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s

How much work is done when 100 N of force is applied to a rock to move it 20 m

Answers

Answer: 2000 J

Explanation: work W = F s

Mechanical energy is the most concentrated form of energy.
a. true
b. false

Answers

( False )Nuclear energy is the most concentrated form of energy.

Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as Carl’s behavior is being influenced by

Answers

He is influenced by EXTRINSIC MOTIVATION

1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock

Answers

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the  water, V, is given as follows;

[tex]V_T[/tex] = [tex]V_r[/tex] + V

[tex]V_T[/tex] = A × h₂

∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³

The total volume, [tex]V_T[/tex] = 120 cm³

The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V

∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, [tex]V_r[/tex] = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

why doesn't a radio operating with two batteries function when one of the batteries is reversed? ​

Answers

Answer:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

Discuss how the pressure cooker is designed to achieve temperatures above 100°C.​

Answers

With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.

At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?

Answers

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

Rhodium is in period 5 of the periodic table. What does this tell you about this element

Answers

Answer:

. It is an extraordinarily rare, silvery-white, hard, corrosion-resistant, and chemically inert transition metal. It is a noble metal and a member of the platinum group.

Explanation:

The tendency for an object to remain at rest in continue in motion is called:

Inertia
Motion
Gravity
Force

Answers

Answer:

A Inertia

Explanation:

A: inertia


Explanation: there isn’t one I just need to put 20 words

How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds

Answers

Answer:

they were fast ⛷⛷

Distance in yards is 30

A student's backpack has a mass of 9.6 kg. The student applies a force of 94.08 N [up] while walking through 1.4 km [E] to get to school. Calculate the work done by the student on the backpack

Answers

The student does zero work on the backpack because the upward force applied by the student is acting perpendicular to the backpack's displacement parallel to the ground.

A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............

Select one:

a.
No force exerted


b.
Less than 2000N


c.
Equal to 200 N


d.
Greater than 200 N​

Answers

Answer:

c.

Equal to 200 N..........

Highest density of electrostatic charges in a metal is found where​

Answers

I don't know the answer but I just want points sorry

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