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:
0.08024 W/m²
Explanation:
P = Power
= Distance = 15 m
= Intensity at 15 m = 0.26 W/m²
= Distance = 27 m
= Intensity at 27 m
Sound intensity is given by
So, we have the relation
The sound intensity at the given distance is 0.08024 W/m²
Answer:
s = 307.34 m
Explanation:
In order to find the distance covered by the dragster during the given time, we will use second equation of motion. The second equation of motion is written as follows:
s = Vi t + (0.5)at²
where,
s = distance covered by the dragster = ?
Vi = Initial Velocity = 0 m/s
t = time interval = 3.97 s
a = acceleration = 39 m/s²
Therefore,
s = (0 m/s)(3.97 s) + (0.5)(39 m/s²)(3.97 s)²
<u>s = 307.34 m</u>
Answer:
Power is energy divided by time taken for it to happen
Explanation:
Energy= 7500J
times taken =2.3s
thus power=7500J÷2.3s
=3260.87(J/s) or 3260.87W
