The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
Explanation :
Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.
Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.
When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.
When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.
The answer is 8 because multiplying 7 and 8 is 56
V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
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