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sveticcg [70]
3 years ago
13

What isFigure ground in psychology

Physics
1 answer:
vivado [14]3 years ago
3 0

Answer:

Figure-ground organization is a tupe of perceptual grouping that is a vital necessity for recognizing objects trough vision.

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How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at
amid [387]
Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
5 0
3 years ago
Why does a comet's tail point away from the Sun? 1Choice 1 The solar wind blows the tail away from the Sun. 2 It is being pulled
AnnyKZ [126]

Choice 1

The Sun's radiation and solar wind cause the dust and gas around the comet (coma) to stretch the coma. The solar wind electromagnetically blows the ions in the coma away.

4 0
3 years ago
In a displacement versus time graph, what does the slope of a line at any point indicate?
Rom4ik [11]

In a displacement versus time graph, the slope of the line at any point on the graph indicates the <em>magnitude of velocity</em>.  

(It can't indicate velocity completely, because the graph shows nothing about the direction of the motion.)

4 0
3 years ago
Read 2 more answers
Select the correct answer.
Juliette [100K]

Answer: the answer is C

Explanation:

4 0
3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
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