Answer:
a - As long as the time between 2 events is reconcilable with a light signal, the time between the events, in that frame, can be determined.
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity 
will be expressed by the following equation:
Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
D.
All follow an elliptical path that has two foci, rather than a circular path.
Kepler found paths are elliptical, not circular
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
