Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V
C1= 3×10^-6 F
V1= 480v
C2= 4×10^-6 F
V2= 500v
(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V
Simplifying the above, we get:
( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.
Further simplified as:
3440 × 10^-6 = 7 × 10^-6 × V
Making V the subject
V = 491.43volts
Therefore the potential difference across each capacitor is 491.43v
Answer:
0.025 m
0.059166 m
Explanation:
P = Pressure
A = Area
x = Compression of spring
Force is given by

From Hooke's law

The spring is compressed 0.025 m
In the second case


Net force would be

Compression would be

The compression of the spring is 0.059166 m
The electrical free path of electron is disturbed with vibrations not normal to normal / room temperature because the electrical heater is agitating the conductor material. this causes the electrons to vibrate more creating more resistance to the flow.
Sound waves are mechanical waves so they need a medium for propagation.
Explanation:
The most fundamental properties of sound waves are:-
1. Wavelength
2. Frequency
3. Amplitude