Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,
Therefore the percent of the atom's volume is occupied by mass is 
Answer:
<span>GPE=81000J or 81kJ</span>
Explanation
Potential Energy = mgh = 20 x 9.8 x ?
<span>To find H use one of the equation of motion </span>
<span>= [(90)^2 - 0 ] / 2(9.8) </span>
<span>Potential Energy = mgh = 20 x 9.8 x 8100 /2(9.8) = 81000 J</span>
Because you have to study about electricity, first of all, and also need to study about polarity
