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2 years ago

Match each type of consequence with its resulting behavior change

Physics

1 answer:

ira [324]2 years ago

5 0

Answer:

Janet stops parking in handicapped spaces after she gets a big parking ticket. - Positive Punishment

Peter’s recess is taken away to discourage him from getting into fights with the other children. - Negative Punishment

Ted increases paying his bills on time to avoid a late fee. - Negative Reinforcement

Sally increases the amount of work she completes to receive more pay. - Positive Reinforcement

Explanation:

In operant conditioning, the main principle is that behavior increases or decreases its frequency depending on whether it's reinforced or punished. A behavior can be reinforced by giving something the subject appreciates, like more pay for their work (positive reinforcement) or taking away something they dislike, like late fees (negative reinforcement). Punishments work the same way, you can give something the subject dislikes, like a parking ticket, (positive punishment) or taking away something they like recess for a child. (negative punishment).

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The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is

kirill [66]

Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
L = k/(d^2)
For Case 1,
L1 = k/(d1^2)
For Case 2,
L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2.
Therefore, the loudness at your friend's position is 16 times that of yours.

5 0

2 years ago

Read 2 more answers

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

2 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

1 year ago

A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material

worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

$r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m$ is the radius

$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

$d=\frac{m}{V}$

where m = 1 kg is the mass. Substituting, we find

$d=\frac{1 kg}{7.1\cdot 10^{-5} m^3}=14,160 kg/m^3$

3 0

2 years ago

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t

arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vector $n=-0.60\hat{i}-0.080\hat{j}$

We need to calculate the torque on the loop

Using formula of torque

$\tau=NIAB\sin\theta$

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

$\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}$

$\tau=4.7\times10^{-4}\ N-m$

Hence, The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

5 0

1 year ago