Answer:
D) Q/2
Explanation:
The relationship between charge Q, capacitance C and voltage drop V across a capacitor is
(1)
In the first part of the problem, we have that the charge stored on the capacitor is Q, when the voltage supplied is V. The capacitance of the parallel-plate capacitor is given by
where is the vacuum permittivity, A is the area of the plates, d is the separation between the plates.
Later, the voltage of the battery is kept constant, V, while the separation between the plates of the capacitor is doubled: . The capacitance becomes
And therefore, the new charge stored on the capacitor will be
Answer:
25,000 joules
given the formula
Q=Mc(final temperature - initial temperature)
M=50 grams convert to kilogram 0.5
c= 4200 j/kg/k
final temperature=32c
initial temperature=20c
<span>A)<span>Constructive and destructive interference of sound waves.</span></span>
Answer:
<em>b. The force is 6 times the original</em>
Explanation:
<u>Coulomb's Law
</u>
The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Written as a formula:
Where:
q1, q2 = the objects' charge
d= The distance between the objects
Suppose one of the charges is doubled, i.e., q1=2q1 and the other charge is tripled (q2'=3q2). The new force F' would be:
Operating:
Substituting the original value of the force:
b. The force is 6 times the original
Answer:
Maybe she kicks it to a wall and it bounces back to her?
Explanation: