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2 years ago

6

What role does a mild electric shock play in Skinner's operant chamber? A. negative reinforcer B. positive reinforcer C. punishe

r D. neutral operant

1 answer:

photoshop1234 [79]2 years ago

8 0

Answer: b

Explanation:

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The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concen

Sever21 [200]

Answer:

0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

The rate of the reaction is about 0.00325 moles/liter/second.

_____

This is the rate of decrease of the concentration of A.

5 0

2 years ago

Read 2 more answers

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal

Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

p = m1 v1 + m2 v2

Part B.

Find the derivative

dp / dt = m1 dv1dt + m2 dv2 / dt

dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1 f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0

2 years ago

Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm

ozzi

Answer:

R = 0.992 Ω

Explanation:

En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.

Matemáticamente,

Resistencia = (resistividad * longitud) / Área De la pregunta;

Resistividad = 3,5 * 10 ^ -5 Ωm

longitud = 170 cm = 1,7 m

Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2

Conectando estos valores a la ecuación anterior, tenemos;

Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =

(3.5 * 1.7) / 6 = 0.992 Ω

3 0

2 years ago

A boy standing on a 19.6 meter tall bridge sees a motorboat approaching the bridge at a constant speed. When the boat is 27 mete

azamat

Answer:

A. 12 m/s

Explanation:

Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.

The equation that describes the motion of the stone is:

y = y_0 + v_0 * t+1/2 * a * t^2

The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:

-19,6 m = -1/2 * 9,8 m/s^2 * t^2

t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2

t=√(4,012 s^2) = 2,003 s

Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:

v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s

Have a nice day! :D

8 0

2 years ago

Transverse waves on a string have wave speed v=8.00 m/s, amplitude A=0.0700m, and direction, and at t=0 the x-0 end of the wavel

Vilka [71]

Answer:

a. frequency = 25 Hz, period = 0.04 s , wave number = 19.63 rad/m

b. y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

c. 0.0496 m

d. 0.03 s

Explanation:

a. Frequency, f = v/λ where v = wave speed = 8.00 m/s and λ = wavelength = 0.320 m

f = v/λ = 8.00 m/s ÷ 0.320 m = 25 Hz

Period, T = 1/f = 1/25 = 0.04 s

Wave number k = 2π/λ = 2π/0.320 m = 19.63 rad-m⁻¹

b. Using y = Asin(kx - ωt) the equation of a wave

where y = displacement of the wave, A = amplitude of wave = 0.0700 m and ω = angular speed of wave = 2π/T = 2π/0.04 s = 157.08 rad/s

Substituting the variables into y, we have

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

c. When x = 0.360 m and t = 0.150 s, we substitute these into y to obtain

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

y = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s × 0.150 s)]

y = (0.0700 m)sin[(7.0668 rad) - (23.562 rad)]

y = (0.0700 m)sin[-16.4952 rad]

y = (0.0700 m) × 0.7084

y = 0.0496 m

d. For the particle at x = 0.360 m to reach its next maximum displacement, y = 0.0700 m at time t. So,

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

0.0700 m = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s)t]

0.0700 m = (0.0700 m)sin[(7.0668 rad - (157.08 rad/s)t]

Dividing through by 0.0700 m, we have

1 = sin[(7.0668 rad - (157.08 rad/s)t]

sin⁻¹(1) = 7.0668 rad - (157.08 rad/s)t

π/2 = 7.0668 rad - (157.08 rad/s)t

π/2 - 7.0668 rad = - (157.08 rad/s)t

-5.496 rad = - (157.08 rad/s)t

t = -5.496 rad/(-157.08 rad/s) = 0.03 s

6 0

2 years ago