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irga5000 [103]
2 years ago
15

You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p

ossible from a conservation of energy standpoint. Also, can you relate this behavior to the transient (natural) response of the circuit that you observed in the previous lab
Physics
1 answer:
nydimaria [60]2 years ago
6 0

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

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550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
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Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
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7 0
2 years ago
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If isomerization requires breaking the pi bond, what minimum energy is required for isomerization in j/mol?
aliina [53]
<span>The minimum energy required for isomerization is 267 000 J/mol 
</span>

The isomerization of cis-but-2-ene to trans-but-2-ene requires breaking of the π bond.

The bond energy of a C-C σ bond is 347 kJ/mol.

The bond energy of a C=C double bond (σ + π) is 614 kJ/mol.

So the bond energy of a π bond is (614 – 347) kJ/mol = 267 kJ/mol =
267 000 J/mol.

6 0
2 years ago
Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
2 years ago
An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass
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I = 7/5*mR^2 M = 7/5*m 
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7 0
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