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1)
The connections between neurons in the retina, specifically the connections referred to as “lateral inhibition,” help us see which of the following better?

A) Contrast
B) Faces
C) Colors

2)
Improving the contrast of an image (making the dark regions darker and the light regions lighter) helps us to identify:

A) The edges of objects
B) The center of objects
C) The color of an object

3)
What assumption does our visual system make in order to see curved surfaces (domes, holes)?

A) Light comes from above
B) Curved surfaces are always evenly lit
C) Curved surfaces are always easy to see, no assumptions are made

4)
Which part of the face does our brain pay the most attention to?

A) Eyes and mouth
B) Eyes and ears
C) Eyes and chin

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
D) All of the above

Hope that helps :)
*the correct answers are bolded, italicized, and underlined.*

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

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With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$