Answer:
a
The speed is
b
The distance is 
Explanation:
From the question we are told that
The speed of the banana is 
The distance from my location is
The time taken is 
The speed of the ice cream is

substituting values


The distance of separation between i and Valdimir is the same as the distance covered by the banana
So
substituting values

Answer: c
Explanation:
The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.
a. 2×2 atoms of Al ≠ 3×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!
b. 3×2 atoms of Al ≠ 3×1 atoms of Al
3×3 atoms of O ≠ 2×2 atoms of O
c. 2×2 atoms of Al = 4×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH ARE EQUAL, CORRECT ANSWER!!!
d. 2×2 atoms of Al ≠ 1×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
Answer:
The magnitude of the second charge is
or 
Explanation:
The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.
The electric potential energy of some charge
at a point in the electric field of another charge
is given by the product of the amount of charge
and electric potential at that point due to the charge
.

The electric potential at that point is given by

where
is the Coulomb's constant.
Therefore,

Now, We have given two charges
and
, whose value is to be found.
When the two charges are infinitely dar apart, the electric potential energy of the system is given by

When the coordinates of position of the two charges are

The distance between the two charges is given by

The electric potential energy of the charges in this configuration is given by

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.
Therefore,

Answer: It prevents us from seeing most of the galactic disk with visible and ultraviolet light.
(hope this helps) :)