The total capacitance is <em>C</em> such that
1/<em>C</em> = 1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)
Solve for <em>C</em> :
<em>C</em> = 1 / (1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)) ≈ 3.1 µF
Answer:
0.05 cm
Explanation:
The compression of the original spring = 12 - 8.55 cm = 3.45 cm = 0.0345 m
By Hooke's law, F = ke
Where F is the applied force, k is the spring constant and e is the extension or compression. In the question, F is the weight of the car.
k = F/e = 1355 × 9.8 / 0.0345 = 384898.55 N/m
This is the spring constant of the original spring. The question mentions that the force constant of the new spring is 5855.00 N/m smaller. Hence, the force constant of the new spring is 384898.55 - 5855 = 379043.55 N/m
With the new spring installed, the compression will be
e = F/k = 1355 × 9.8 / 379043.55 = 0.035 m = 3.5 cm
The difference in the compressions of both springs = 3.5 cm - 3.45 cm = 0.05 cm
Answer:
Explanation: Is their more?
Initial conditions
K is the spring constant
X is the compression of the spring