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3 years ago

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In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the cir

cuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.

1 answer:

Ahat [919]3 years ago

4 0

Answer:

The energy is $U = 18.98 \ J$

Explanation:

From the question we are told that

The inductor is $L = 4.74 \ H$

The resistance of the resistor is $R = 9.33 \ \Omega$

The voltage of the battery is $V = 26.4 \ V$

Generally the current flowing in the circuit is mathematically represented as

$I = \frac{V}{R}$

=> $I = \frac{26.4}{9.33 }$

=> $I = 2.83 \ A$

Generally the corresponding energy stored in the circuit is

$U = \frac{1}{2} * L * I^2$

$U = \frac{1}{2} * 4.74 * 2.83 ^2$

$U = 18.98 \ J$

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When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

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So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to which the body may safe.

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A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

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Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

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<h3>What is the Long straight distance?</h3>

The line that runs form one end of the circle to another is called the diameter of the circle. The pool is a circle according to the question and the long straight distance that a person can swim is the same of the diameter of the circular pool.

Now we have;

A = πr^2

A = area of pool

r = radius of pool

r = √A/ π

r = √25/3.142

r = 2.82m

Diameter of the circular pool = 2 r = 2 (2.82 cm) = 5.64 m

Learn more about circle: brainly.com/question/11833983

#SPJ1

Missing parts;

An ad for an above-ground pool states that it is 25 m2. From the ad, you can tell that the pool is a circle. If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swim? Express your answer in three significant figures.

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