Answer:
No, not necessarily
Explanation:
If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.
An example of an object that has zero velocity but non-zero acceleration:
If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 
rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 .
Answer:
y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
1 / i = 1 / f - 1 / o
1 / i_red = 1 / f_red - 1 / o
1 / i_red = 1 / 19.57 - 1/30
1 / i_red = 1,776 10-2
i_red = 56.29 cm
Blue light
1 / i_blue = 1 / f_blue - 1 / o
1 / i_blue = 1 / 18.87 - 1/30
1 / i_blue = 1,966 10-2
i_blue = 50.863 cm
Now let's use the magnification ratio
m = y ’/ h = - i / o
y ’= - h i / o
Red Light
y_red ’= - 5 56.29 / 30
y_red ’= - 9.3816 cm
Light blue
y_blue ’= 5 50,863 / 30
y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
y_red ’/ y_blue’ = 9.3816 / 8.47716
y_red / y_blue = 1,107
y_red / y_blue = 1.11
Average speed = distance travelled / time used
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1. A. 6.00 sec
The graph shows the velocity of an object (y-axis) versus the time (x-axis). In order to find when the magnitude of the velocity reaches 36.00 km/h, we should find the time t (x-coordinate) at which the velocity (y-coordinate) is 36.
By looking at the graph, we see that this occurs when t=6.00 s.
2. A. positive acceleration
In a velocity-time graph like this one, the slope of the curve corresponds to the acceleration of the object. In fact, acceleration is defined as:
where 
is the variation of velocity and 
is the variation of time. We see that this quantity corresponds to the slope of the curve in the graph (in fact, represents the increment of the y coordinate, while represents the increment of the x coordinate). So, a positive slope means a positive acceleration: in this case, the slope is positive, so the acceleration is also positive.
