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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The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.
<h3>What mass of lead should be placed on the cube?</h3>
Given: Side of the cube (a) = 20cm
The density of the cube (ρc) =
a) Applying the force balance, the buoyant force must be equal to the weight of the cube
ρcgV = ρg × (Ax)
Substituting the values in the above equation, we get
x = 0.13
where x is the height of the cube in the water
is the area of the cross-section
ρ is the density of the water
V is the volume of the cube
Now, the height above the surface of the water would be
h = a − x
Substituting the values, then we get
h = 0.2 − 0.13
h = 0.07 m
b) The mass added is "m" so the complete cube is submerged in the water, therefore
ρcgV + mg = ρg × (V)
m = 2.8 kg
The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.
To learn more about buoyant force refer to:
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