Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.
Kinetic energy = (1/2) (mass) (speed)²
The rock's kinetic energy is not
(1/2) (4 kg) (10 m/s)²
= (1/2) (4 kg) (100 m²/s²)
= 200 Joules .
It may be more, or it may be less. The only thing
we can be sure of is that it is not 200 Joules.
Hope this helps mark brainest plz
Answer:
The distance between the two spheres is 914.41 X 10³ m
Explanation:
Given;
4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;
1 e = 1.602 X 10⁻¹⁹ C
4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C
V = Ed
where;
V is the electrical potential energy between two spheres, J
E is the electric field potential between the two spheres N/C
d is the distance between two charged bodies, m

where;
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063
d = 914.41 X 10³ m
Therefore, the distance between the two spheres is 914.41 X 10³ m