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3 years ago

6

A 10-μF capacitor in an LC circuit made entirely of superconducting materials ( R = 0 Ω ) is charged to 100 μC. Then a supercond

ucting switch is closed. At t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. At a later time, Vab = +10V . At that time, Vdc is

1 answer:

satela [25.4K]3 years ago

8 0

Answer:

Vdc=10V

Explanation:

in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,

if the loop remains closed for a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

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Que fuerza se obtendrá en el émbolo mayor de una prensa hidraúlica cuya área es de 167 cm2, cuando el émbolo menor de área igual

stiv31 [10]

Responder:

<h2>5.368N, </h2>

Explicación:

Según el principio pascal, establece que la presión aplicada en un punto sobre un líquido en un recipiente cerrado es igual a igual a la presión en cualquier otro punto del líquido.

Matemáticamente Presión ejercida por el pistón más pequeño = Presión ejercida por el pistón más grande.

La presión es la relación entre la fuerza y su área de sección transversal.

P = Fuerza / Área de sección transversal

Sea P1 la presión sobre el pistón más pequeño y P2 la presión ejercida por el pistón más grande.

Como P1 = P2 entonces;

F1 / A1 = F2 / A2

Dado F1 = 450N, A1 = 14 cm², A2 = 167 cm² y F2 =?

Sustituyendo el valor conocido en la fórmula para obtener el requerido, tenemos;

$\frac{450}{14} = \frac{F2}{167}\\ \\Cross\ multiplying\\\\14*F_2 = 450*167\\\\14F_2 = 75,150\\\\F_2 = \frac{75,150}{14} \\\\F_2 = 5,367.9N$

$F_2 \approx 5, 368N$

Por lo tanto, la fuerza que se obtendrá en el pistón más grande de una prensa hidráulica cuya área es de 167 cm² es aproximadamente 5,368N,

4 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

Please help me with this question...

vredina [299]

The answer is B (The second one). I'm not sure though.

3 0

3 years ago

The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration

a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>

Forces along x axis direction are as follows

4N along +x axis, so it's taken as +4 N
2N along -x axis , so it's taken as -2N.

Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>

As per Newton's second law, Force = mass × acceleration of the object
Force along x axis= mass × acceleration along x axis= 2N
Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

3 0

2 years ago

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

Oxana [17]

Answer:

A) $I_{total}$ = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = $I_{cm}$ + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

$I_{total}$ = $I_{body}$ + 2 $I_{arm}$

$I_{body}$ = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

$I_{arm}$ = M D²

Let's write the equation

$I_{total}$ = ½ M R² + 2 (m D²)

Let's calculate

$I_{total}$ = ½ 56 0.20² + 2 4 0.20²

$I_{total}$ = 1.12 + 0.32

$I_{total}$ = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0

4 years ago