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3 years ago

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Physics

1 answer:

inessss [21]3 years ago

8 0

This is an example of classical conditioning, and the library is a conditioned stimulus.

The idea of classical conditioning was first put forward by the man Ivan Pavlov, a Russian physiologist.

In his studies, he was able to successfully associate a once neutral stimulus to a conditioned response. The neutral stimulus hitherto does not produce the expected response until after conditioning has taken place.

In this case, the library was a neutral stimulus. When it becomes associated with a bakery, it provokes the conditioned response of hunger any time you go to the library.

The library is now a conditioned stimulus.

Learn more: brainly.com/question/18080270

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What type of system would allow light and air to enter and exit? A. Connected

Svetradugi [14.3K]

Answer:

An open system.

Explanation:

An isolated system allows the exchange of neither energy nor matter with the surroundings. A closed system allows the exchange of energy, but not matter. An open system allows the exchange of both energy and matter.

Notice that in this question, light (electromagnetic wave) is a form of energy. The entry and exit of light allows this system to exchange energy with its surroundings- just as how the earth receives energy from the sun. Additionally, this system could exchange energy with its surroundings through the exchange of matter (in particular, air) with its surroundings.

Thus, the system in this question is an open system.

8 0

2 years ago

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Oh football player kicks a football from the height of 4 feet with an initial vertical velocity of 64 ft./s use the vertical mot

Stolb23 [73]

Answer:

4.1 seconds

Explanation:

The height of the football is given by the equation:

$H = -16t^2 + V*t + S$

Using the inicial position S = 4 and the inicial velocity V = 64, we can find the time when the football hits the ground (H = 0):

$0 = -16t^2 + 64*t + 4$

$4t^2 - 16t - 1 = 0$

Using Bhaskara's formula, we have:

$\Delta = b^2 - 4ac = (-16)^2 - 4*4*(-1) = 272$

$t_1 = (-b + \sqrt{\Delta})/2a$

$t_1 = (16 + 16.49)/8 = 4.06\ seconds$

$t_2 = (-b - \sqrt{\Delta})/2a$

$t_2 = (16 - 16.49)/8 = -0.06\ seconds$

A negative time is not a valid result for this problem, so the amount of time the football is in the air before hitting the ground is 4.1 seconds.

7 0

3 years ago

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A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100

Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

$\rho$ = Density of fluid

v = Fluid velocity

m = Mass = $\rho Al$

Centripetal force is given by

$F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L$

Pressure is given by

$P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s$

The angular speed of the tube is 31.321 rad/s

5 0

3 years ago

Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200

artcher [175]

We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

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3 years ago

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In your research lab, a very thin, flat piece of glass with refractive index 1.20 and uniform thickness covers the opening of a

liraira [26]

Answer:

(a). The thickness of the glass is 868 nm.

(b). The wavelength is 3472 nm.

Explanation:

Given that,

Refractive index = 1.20

Wavelength = 496 nm