The equivalent capacitance between A and B points is 2.5F.
<h3>What is parallel plate capacitor?</h3>
The two parallel plates placed at a distance apart used to store charge when electric supply is on.
The capacitance of a capacitor is given by
C = ε₀ A/d
From the given circuit C1, C2 and C3, C4 are in parallel C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F
C1, C2 = 4 +4 =8F
C3, C4 = 2 +4 =6F
Now , all capacitors are in series.
Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2
Ceq = 2.5 F
Thus, the equivalent capacitance between A and B points is 2.5F.
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Answer:
if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Explanation:
the average velocity of the car is 15 m/s example I have this on a test
