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Softa [21]
2 years ago
6

Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.

Physics
2 answers:
Naddika [18.5K]2 years ago
6 0

One area in which cognitive psychologists have influenced traditional conditioning practices is modern education. Education previously employed conditioning models to teach students with reward (good scores) and punishment (bad scores). Modern educators use metacognition (a concept from cognitive psychology) to aid students in evaluating their personal knowledge. That in turn helps students to strategize for improvement in school subjects where they are weak.


user100 [1]2 years ago
4 0
Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.
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Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -2.0 µC; sphere B carries a charge of -6.0 µC;
Dmitry_Shevchenko [17]

Answer:

None of the option is correct

A=-4µC, B=0 C, C=0 C

C will be +4.0 µC deficient after the contact

Explanation:

A and B are come in contact together, the charge will flow to establish  equilibrium, and hence becoming: A=-4µC, B=-4µC, C=+4.0 µC

Similarly when C and B touch, the positive and the negative will exact the same force due to equal charge magnitude and become electrically neutral : A=-4µC, B=0 C, C=0 C.

5 0
2 years ago
A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

Answer:

 Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

Explanation:

  Force applied on baseball = 30 times weight of the ball.

   Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

                                 a = 30g

   So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

8 0
2 years ago
The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un
Elza [17]

Answer:

The answer to your question is: 15 m/s2

Explanation:

Equation    x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find  x at t = 4.1 s

x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2

6 0
2 years ago
One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has
vovangra [49]

Answer:

The mass of the cube is 420.8 kg.

Explanation:

Given that,

Length of edge = 38.9 cm

Density \rho= 7.15 \times10^{3}\ kg/m^3

We need to calculate the volume of cube

Using formula of volume

V = 38.9^3

V=0.058863\ m^3

We need to calculate the mass of the cube

Using formula of density

\rho = \dfrac{m}{V}

m = V\times\rho

m =0.058863\times7.15 \times10^{3}

m=420.8\ kg

Hence, The mass of the cube is 420.8 kg.

7 0
2 years ago
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

H_{nom}=2hp=1491.4W

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s

F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N

(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N

T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N

F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N

b)

H_a=1491*1.25=1863.75W

n_f_s=\frac{H_a}{H_{nom}K_S }=1

dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

7 0
2 years ago
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