<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
Answer:
Hydraulic pressure exerted on glass slab, ρ=10 atm
Bulk modulus of glass, B=37×10^9 Nm^−2
Bulk modulus, B=P/(ΔV/V)
where,
ΔV/V= Fractional change in volume
ΔV/V=P/B
=10×1.013×10^5 /(37×10 ^9)
=2.73×10^-5
Therefore, the fractional change in the volume of the glass slab is 2.73×10^-5
Hope it helps
work done=446.9 J . so option (c) is correct.
Explanation:
the formula for work done is given by
W= F d
F= force= mg where m= mass and g= acceleration due to gravity
F= 3.8 (9.8)=37.24 J
so W=37.24 (12)
W=446.9 J
Y = +1-3 = -2
X = -5+7 = +2
D = √2^2-2^2 = 2√1+1 = 2√2 km
Using Newton's Second Law, F = ma, where F is the net force
So the net force is:
F = (6kg)(4m/s^2) = 24N
Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.
30N-24N = 6N
