Answer:
23.52 m/s
Explanation:
The following data were obtained from the question:
Time taken (t) to reach the maximum height = 2.4 s
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) =..?
At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)
was thrown as follow:
v = u – gt (since the rock is going against gravity)
0 = u – (9.8 × 2.4)
0 = u – 23.52
Collect like terms
0 + 23.52 = u
u = 23.52 m/s
Therefore, the rock was thrown at a velocity of 23.52 m/s.
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
Answer:
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