The energy of a certain charged capacitor is 7 J. What is the new energy stored in that capacitor if its charge is decreased to
1 answer:
Answer:
The new energy is 1/16 of the original energy
Explanation:
A capacitor is an electric device which is able to store charge when it is connected to a power supply.
The energy stored in a capacitor is given by the equation
where:
Q is the charge stored on the capacitor
C is the capacitance of the capacitor
For the capacitor in this problem, initially we have
E = 7 J (energy) when the charge stored is Q and the capacitance is C
Later, the charge is decreased to 1/4 of its original value, so the new charge is
Since the capacitance remains the same, the new energy is
Therefore, the new energy is 1/16 of the original energy.
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The gravitational force between two masses m₁ and m₂ is
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
(1)
There are only two forces acting on the book:
- its weight, directed downward: mg
- the force exerted by the hand on the book, of 20 N, directed upward
so, equation (1) becomes
from which we can calculate the book's acceleration, a:
There are only two forces acting on the book:
- its weight, directed downward: mg
- the force exerted by the hand on the book, of 20 N, directed upward
so, equation (1) becomes
from which we can calculate the book's acceleration, a:
The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge:
, and Vf and Vi are the final and initial voltages.
Substituting the numbers, we find:
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge:
Substituting the numbers, we find: