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Andrew [12]
2 years ago
11

In operant conditioning, generalization occurs when

Physics
1 answer:
Vitek1552 [10]2 years ago
5 0
Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
Hope this helps!
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Which is usually greater for a pair of surfaces, the coefficient of static friction or the coefficient of sliding friction? why
lorasvet [3.4K]

Explanation :

Static friction is the frictional force between two objects that are at rest. While sliding friction is the frictional force between two objects in contact and are sliding w.r.t each other.

Static friction is usually greater than sliding friction because in static friction the contact forces is more and the interlocking between objects is tight as compared to sliding friction.    

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How do scientists determine the age of the solar system?
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The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

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What is an output force?
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The answer would be D

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