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sergiy2304 [10]
2 years ago
10

Dr. Clairmont will have his psychology students training dogs from the local animal shelter to sit using various types of reinfo

rcements. One group of students will train the dogs
to sit by giving the dog a treat when it correctly performs. The other group of students will train the dogs to sit by firmly and loudly scolding the dog when it does not correctly
perform. After 3 weeks of training, the dogs will be evaluated to see which group has more dogs that will sit on command.
What might an ethics review committee decide about the proposal?
The proposal is approved, as it meets the ethical guidelines related to animal research.
The proposal is approved, because while the animals suffer harm, the human participants suffer no psychological harm.
The proposal is denied, because the animal participants suffer physical harm.
The proposal is denied, because the animal participants cannot leave the study
The proposal is denied, because the human participants are deceived
Physics
1 answer:
olga2289 [7]2 years ago
6 0

Answer:Habituation is a simple learned behavior in which an animal gradually stops responding to a repeated stimulus.

Imprinting is a specialized form of learning that occurs during a brief period in young animals—e.g., ducks imprinting on their mother.

In classical conditioning, a new stimulus is associated with a pre-existing response through repeated pairing of new and previously known stimuli.

In operant conditioning, an animal learns to perform a behavior more or less frequently through a reward or punishment that follows the behavior.

Some animals, especially primates, are capable of more complex forms of learning, such as problem-solving and the construction of mental maps.

Introduction

If you own a dog—or have a friend who owns a dog—you probably know that dogs can be trained to do things like sit, beg, roll over, and play dead. These are examples of learned behaviors, and dogs can be capable of significant learning. By some estimates, a very clever dog has cognitive abilities on par with a two-and-a-half-year-old human!

Explanation:

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Answer:

 V₁ = 6 V ,  V₂ = V₃ = 3 V

Explanation:

To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.

* a serial circuit there is only one path for current

in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term

* Parallel circuit in this there are two or more paths for the current

in this circuit the voltage is constant and the east is divided between each branch

with these principles let's analyze the proposed circuit

The DC battery is in parallel with resistor R1 and the equivalent of the other branch,

as in a parallel circuit the voltage is constant

               V₁ = 6 V

in the other branch (23) it forms a series construction, where the current is constant

               6 = iR₂ + iR₃

as they indicate that each resistance has the same value

              6 = 2 iR

              V = V₂ = V₃ = 3 V

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Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
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