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:


Explanation:
Wave 1, 
Wave 2, 
Wave 3, 
Wave 4, 
The general equation of travelling wave is given by :

The value of
will remain the same if we take phase difference into account.
For first wave,



For second wave,



For the third wave,



For the fourth wave,



It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.
Answer:
2.11 seconds
Explanation:
We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):

Answer:
square metre is the answer
Explanation:
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