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Rina8888 [55]
3 years ago
14

Classical conditioning requires _____.

Physics
2 answers:
LekaFEV [45]3 years ago
5 0
I would say A classical conditioning definitely demands a pairing of two stimuli
hope i helped 
sveta [45]3 years ago
3 0

Answer:

the pairing of two stimuli

Explanation:

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Georgia [21]

Explanation:

7.9x10^9 km is equal to

=7900000000km

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3 years ago
If an object moves in uniform circular motion in a circle of radius R = 1.0 meter, and the object takes 4.0 seconds to complete
Masteriza [31]
A = ω²r

r = 1 m
ω = 2πf = 2π * 10 / 4 = π*5 s⁻¹
4 0
3 years ago
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A concave lens has a focal length of 25cm. it's power in diaptor is​
IgorLugansk [536]

As we know that :

\begin{gathered}\large{\boxed{\sf{P \: = \: \dfrac{1}{f}}}} \\ \\ \rightarrow {\sf{P \: = \: \dfrac{1}{-25}}}\end{gathered}

Power, is in Meter. So divide focal length by 100

\begin{gathered}\rightarrow {\sf{P \: = \: \dfrac{1}{\dfrac{-25}{100}}}} \\ \\ \rightarrow {\sf{P \: = \: \dfrac{-100}{25}}} \\ \\ \rightarrow {\sf{P \: = \:- 4}} \\ \\ \underline{\sf{\therefore \: Power \: of \: Concave \: lens \: is \: - \: 4D}}\end{gathered}

8 0
2 years ago
The average density of living matter on earthâs land areas is 0.10 g/cm2 . what mass of living matter in kilograms would occupy
Margaret [11]
<span>Density is a value for mass, such as kg, divided by a value for volume or area. Density is a physical property of a substance that represents the mass of a certain substance per unit volume or unit area. From the problem statement, we are given the density of the living matter in a particular area and the area. To calculate mass, we simply multiply the area to the density. However, we should remember that the units should be homogeneous or the same and if not, we convert.

Mass = density x area
Mass = 0.10 g / cm^2 (0.125 ha) ( 1x10^8 cm^2 / 1 ha )
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4 0
3 years ago
Suppose you place an object 8 cm in front of a converging lens and the image appears 16 cm on the other side of the lens. What i
Marianna [84]

Answer:

5.33 cm

Explanation:

The lens equation states that:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem,

p = 8 cm

q = 16 cm ( the sign is positive since the image is real, which means it is formed on the other side of the lens)

Substituting into the equation,

\frac{1}{f}=\frac{1}{8 cm}+\frac{1}{16 cm}=\frac{3}{16 cm}

f=\frac{16}{3}cm=5.33 cm

3 0
3 years ago
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