Half the potential difference of the the1-µF
A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.
The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.
Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.
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Answer:
The jp2003parker guy is extremely wrong
So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.
Explanation:
Answer:
D. Histogram.
Explanation:
A histogram with equal intervals is suitable here.
Answer:
If one side of the train is positive and the other is negative they will attract if they are the same then they will repel.
Explanation:
If both are positive they will repel if both are negative they will repel and if they are opposites they will attract.
Answer:
a)-1.014x 
J
b)3.296 x J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x 
x 47 x 110/ 3.4 => -1.014x J
b) at d= 0.8m (3.4-2.6) and =-1.014x J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+ = 
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore is zero
= +
The final potential energy is
= - GMaMb/d
Solving for ' '
= + GMaMb/d => -1.014x + 6.67 x x 47 x 110/ 0.8
= 3.296 x J
