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3 years ago

Match each type of consequence with its resulting behavior change

Physics

1 answer:

ira [324]3 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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A car is moving at 6 m/s and then accelerates at 1.7 m/s2 for 4.2 seconds. What is the final velocity of the car?

Bad White [126]

Explanation:

Hey there!!

Here,

Initial velocity (u) = 6m/s.

Acceleration (a) = 1.7m/s^2.

Time (t) = 4.2s.

final velocity (v) = ?

We have,

$a = \frac{v - u}{t}$

Putting their values,

$1.7 = \frac{v - 6}{4.2}$

7.14 = v - 6

v = 7.14 + 6

Therefore, the final velocity is 13.14 m/s.

Hope it helps....

5 0

3 years ago

When the force acting on the body equal to acceleration?

topjm [15]

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

5 0

2 years ago

Kamila [148]

Answer: (x - 4)² + (y - 7)² = 9

Explanation:

The equation of a circle is: (x - h)² + (y - k)² = r² where

(h, k) is the center
r is the radius

Given: (h, k) = (4, 7)

Find the intersection of the given equation and the perpendicular passing through (4, 7).

3x - 4y = -1

-4y = -3x - 1

$y=\dfrac{3}{4}x-1$

$m=\dfrac{3}{4}$ --> $m_{\perp}=-\dfrac{4}{3}$

$y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}$

Use substitution to find the point of intersection:

$x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6$

Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius

$r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3$

Input h = 4, k = 7, and r = 3 into the circle equation:

(x - 4)² + (y - 7)² = 3²

(x - 4)² + (y - 7)² = 9

4 0

3 years ago

Math the following to definitions

Leona [35]

Hi,
The answer should be:
Vector 5
Magnatude 4
Weight 1
Mass 2
Net force 6
Force 3

Hope this helps!

6 0

3 years ago

A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric

Tcecarenko [31]

Answer:

a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

Fe + Fm = m a

a = (Fe + Fm) / m

the electric force is

Fe = q E k ^

Fe = 4.25 10-4 60 k ^

Fe = 2.55 10-2 k ^

the magnetic force is

Fm = q v x B

Fm = 4.25 10⁻⁴ $\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]$

fm = 4.25 10⁻⁴ (-j ^ 30 4)

fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

aₓ = 0

there is no force

Axis y

ay = -5.10 10²/5 10⁻⁵ j ^

ay = -1.02 107 j ^ / s2

z axis

az = 2.55 10⁻² / 5 10⁻⁵ k ^

az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

x = vo t + ½ aₓ t₂2

x = 30 0.02 + 0

x = 0.6m

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

y = I + go t + ½ ay t²

y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

y = 1 - 2.04 10³

y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

z = zo + v₀ t + ½ az t²

z = 0 + 0 + ½ 5.1 10² 0.02²

z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

since there is no acceleration the speed remains constant

vₓ = 30.0 m / s

Axis y

let's use the equation v = v₀ + $a_{y}$ t

$v_{y}$ = 0 + -1.02 10⁷ 0.02

v_{y} = 2.04 10⁵ m / s

z axis

v_{z} = vo + az t

v_{z} = 0 + 5.1 10² 0.02

v_{z} = 1.02 10⁻¹m / s

8 0

4 years ago