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.
Answer:
B electrons protons and neutrons
hope i helped...
Explanation:
It’s due to the refraction of light rays as they traverse the boundary between water and air. The greater the viewing angle, the more the bending of the light and therefore the less the apparent depth. An object placed in a denser medium, when viewed from rarer medium appears to be at a lesser depth than its real depth due to refraction of light.
Answer:
See explanations below
Explanation:
According to Newtons second law of motion
F = mass * acceleration
F = ma
If mass of an object is decreased to half, then m₂ = 1/2 m
If acting force is reduced by quarter, then F₂ = 3/4 F
F₂ = m₂a₂
3/4F = 1/2m a₂
Divide both expressions
(3/4F)/F = (1/2m)a₂/ma
3/4 = 1/2a₂/a
3/4 = a₂/2a
4a₂ = 6a
2a₂ = 3a
a₂ = 3/2 a
Hence the acceleration of its motion will be one and a half of its original acceleration.
here we have to calculate the net positive charge present on he surface of the conducting sphere.
as the sphere is a conducting one the,the charge will be stored on its surface.
though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.
the electric field at outside of the sphere is E=
as E= 
so V=
here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .
hence we have 27=
q=27×
coulomb
=3.6×
coulomb [ans]
