Answer:

Explanation:
Given that,
Capacitance 1, 
Capacitance 2, 
Capacitance 3, 
C₁ and C₂ are connected in series. Their equivalent is given by :



Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :



So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
Answer:


Explanation:
Specific Volume 
Absolute Pressure (a) 
Giving



(b) 
Giving



(a)
Generally the equation for quality of Steam X is mathematically given by



(b)
Generally the equation for quality of Steam X is mathematically given by



Initially, the velocity vector is
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by
, so the velocity is
.
Converting back to direction and magnitude, we get 
Answer:
Explanation:
Given that
F=2x³
Work is given as
The range of x is from x=0 to x=D
W=-∫f(x)dx
Then,
W=-∫2x³dx from x=0 to x=D
W=- 2x⁴/4 from x=0 to x=D
W=-2(D⁴/4-0/4)
W=-D⁴/2
W=1/2D⁴
The correct answer is F