An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.
1 answer:
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
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10.3
Explanation:
Step 1:
The pressure exerted by any liquid column of height, h density d is given by the formula P = h * d * g
Step 2:
It is given that one atmosphere pressure pushes up 76.0 cm of mercury, we need to calculate the level of water that will be pushed by the same pressure.
Step 3:
Since the pressure pushing up mercury and water is the same
*
* g =
*
* g
=
= (13.6 g/cm * 76 cm)/1 g/cm = 1033.6 cm
Step 4:
Now we need to express the answer in meters.
1 m = 100 cm.
1033.6 cm = 10.336 m
This can be rounded off to 10.3 m