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Andreas93
11 days 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:
Maru [2.3K]11 days ago
6 0

Answer:

w = √ 1 / CL

This scenario does not breach the principle of energy conservation since the power source's voltage matches the resistance's voltage drop.

Explanation:

This issue pertains to electrical circuits, specifically series RLC circuits, where the resistor, capacitor, and inductor are arranged in series.

In these types of circuits, impedance can be calculated as follows:

X = √ (R² +  (X_{C} -X_{L})² )

Where Xc and XL denote capacitive and inductive impedance, respectively.

X_{C} = 1 / wC

X_{L} = wL

The resonance frequency condition

X_{C} = X_{L}

results in minimal circuit impedance, which maximizes both current and voltage, leading to an observable increase in signal strength.

This phenomenon does not violate energy conservation, as the power source voltage equals the voltage drop across the resistance:

V = IR

Since the impacts of the other two components are neutralized, this occurs for

X_{C} = X_{L}

1 / wC = w L

w = √ 1 / CL

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Keith_Richards [2256]
through the Doppler effect. The formula for apparent frequency is derived as F apparent = F real x (Vair ± Vobserver) / (Vair ± Vsource). In this scenario, should the observer move towards the source—place a positive sign in the numerator and a negative in the denominator. Since the observer approaches the wall, we apply the formula to derive the necessary speed.
4 0
8 days ago
Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
Maru [2355]

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

7 0
29 days ago
A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
Sav [2226]

Answer:

The water level increases more when the cube is above the raft before it sinks.

Explanation:

The principle involved is based on Archimedes' theory, which states that immersing an object in water will raise the initial water level. This means that the height of the water in the container increases. The increase in water volume corresponds to the volume of the submerged object.

We can consider two scenarios: when the steel cube rests on the raft and when it settles at the pool's bottom.

When the cube rests at the pool’s bottom, the volume will indeed rise, and we can ascertain this using the cube's volume.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

When an object floats, it's because the densities of the object and water are in equilibrium.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

The formula for density is:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyant force can be calculated with this equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc indicates that the combined volume of the raft and the cube exceeds that of the cube alone resting at the bottom of the pool. Hence, when the cube is positioned above the raft, the water level rises more before it becomes submerged.

7 0
1 month ago
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i
Softa [2029]

Esta pregunta está incompleta, la pregunta completa es;

La figura muestra un recipiente sellado en la parte superior por un pistón móvil. Dentro del recipiente hay un gas ideal a 1.00 atm. 20.0°C y 1.00 L.

"¿Cuál será la presión dentro del recipiente si el pistón se mueve hasta la marca de 1.60 L mientras se mantiene constante la temperatura del gas?"

Answer:

la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L manteniendo constante la temperatura del gas

Explicación:

Dado que;

P₁ = 1.00 atm

P₂ =?

V₁ = 1 L

V₂ = 1.60 L

la temperatura del gas se mantiene constante

sabemos que;

P₁V₁ = P₂V₂

por lo que hacemos la sustitución

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Por lo tanto, la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L mientras la temperatura del gas se mantiene constante

5 0
27 days ago
. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it
Keith_Richards [2256]

Ans    specifically, 4

Explanation:

7 0
15 days ago
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