Answer:
5702.88 J or 5.7mJ
Explanation:
Given that :
C 1 = 6.0-μF
C 2 = 4.0-μF
V 1 = 50V
V 2 = 34V
Note that : Q = CV
Q 1 = C1 * V1
Q 1 = 50×6 = 300μC
Q 2 = 34×4 = 136μC
Parallel connection = C 1 + C 2
= 6+4 = 10μC
V = Qt/C
Where Qt = Q1+Q2
V = Q1+Q2/C
V = 300+136/10
V = 437/10
V = 43.6volts
Uc1 = 1/2×C1V^2
= 1/2 × 6μF × 43.6^2
= 1/2 × 6μF × 1900.96
= 3μF × 1900.96volts
= 5702.88J
= 5702.88J/1000
= 5.7mJ
Answer: a) 95.07m b) 81.88 m
Explanation:
a)
For finding the distance when vehicle is going downhill we have the formula as:
Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)
Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31
Reaction time= 0.28
So putting values we get
Stop sight distance= 0.28*72.4 *1 + 
Stop sight distance= 95.07 m
b)
For finding the distance when vehicle is going uphill we have the formula as:
Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)
Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31
Reaction time= 0.28
So putting values we get
Stop sight distance= 0.28*72.4 *1 + 
Stop sight distance= 81.88 m
It forms mountains and sometimes islands.