1)
The connections between neurons in the retina, specifically the connections referred to as “lateral inhibition,” help us see which of the following better?
<em><u>A) Contrast</u></em>
B) Faces
<span>C) Colors
2)
</span>Improving the contrast of an image (making the dark regions darker and the light regions lighter) helps us to identify:
<em><u>A) The edges of objects</u></em>
B) The center of objects
<span>C) The color of an object
</span>
3)
What assumption does our visual system make in order to see curved surfaces (domes, holes)?
<em><u>A) Light comes from above</u></em>
B) Curved surfaces are always evenly lit
<span>C) Curved surfaces are always easy to see, no assumptions are made
</span>
4)
Which part of the face does our brain pay the most attention to?
<u><em>A) Eyes and mouth</em></u>
B) Eyes and ears
<span>C) Eyes and chin
</span>
5)
If all these assumptions sometimes lead to mistakes, for example in these optical illusions, why do we make them?
A) It helps us see things faster
B) It helps us see things correctly
C) It helps us pay attention to what's important
<span><em><u>D) All of the above
</u></em></span>
Hope that helps :)
*the correct answers are bolded, italicized, and underlined.*
Answer: of gamma rays with high energies.
Explanation:Gamma-ray bursts are extremely energetic explosions that have been observed in distant galaxies. They are the brightest electromagnetic events known to occur in the universe.
The frequency of the most energetic bursts has been measured at around 3.0×1021 3.0 × 10 21 Hz.
Answer:
a. 5.23 m/s² b. 44.23 N
Explanation:
a. What is the centripetal acceleration of the hammer?
The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.
Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.
So a = rω²
= 1.4 m × (3.74 rad/s)²
= 5.23 m/s²
b. What is the tension in the chain?
The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.
So, T = 8.45 kg × 5.23 m/s²
= 44.23 N
Answer:
This same Hawaii telescope, which would be 4 km across water level, can't provide an appropriate version of distanced planetary bodies. A further overview is provided below.
Explanation:
- The surface area of that same earth's orbit seems to be approximately 480 km heavy. The atmosphere isn't translucent to the only certain wavelength range of the radioactivity. Not because all-stars, as well as gliders, emit specific wavelengths, but several of them generate ultraviolet as well as infrared.
- Those same radiations have either been mediated primarily as well as passes through the atmosphere. Due to the Blockage, they can't even be interpreted with such a similar quality unless the telescope would be positioned throughout the portion of the atmosphere.
What is missing is the density of salt water. Let's denote it ρ.
<span>The pressure at depth d then is: </span>
<span>p = p0 + ρ g d </span>
<span>, where p0 is the atmospheric pressure at the surface of the sea. We assume it is equal to the 1 atm inside the sub. </span>
<span>Hence the pressure difference between inside and outside of the window is: </span>
<span>Δp = ρ g d </span>
<span>The force on the window is this pressure difference, multiplied by the area of the circular window of diameter D. The latter equals </span>
<span>A = ¼ π D². </span>
<span>So </span>
<span>F = p A = ¼ π D ²ρ g d. </span>
<span>This should be less than Fmax. </span>
<span>Hence d < Fmax / (¼ π D ²ρ g ). </span>
