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Svetradugi [14.3K]

4 years ago

6

An inductor is connected to a 18 kHz oscillator. The peak current is 70 mA when the rms voltage is 5.4 V What is the value of th

e inductance L

1 answer:

kobusy [5.1K]4 years ago

5 0

Answer:

The value of the inductance is 0.955 mH

Explanation:

Given;

frequency of the oscillator, f = 18 kHz = 18,000 Hz

the peak current, I₀ = 70 mA = 0.07 A

the root mean square voltage, $V_{rms}$ = 5.4 V

The root mean square current is given as;

$I_{rms}= \frac{I_o}{\sqrt{2} }$

$I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A$

Inductive reactance is given by;

$X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms$

Inductance is given by;

$L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H$

L = 0.955 mH

Therefore, the value of the inductance is 0.955 mH

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An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

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Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

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Answer : The energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

Explanation :

First we have to calculate the mass defect $(\Delta m)$ .

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$^{232}Th\rightarrow ^{228}Ra+^{4}He$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})$

$\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg$

conversion used : $(1amu=1.66\times 10^{-27}kg)$

Now we have to calculate the energy released.

$Energy=\Delta m\times (c)^2$

$Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=7.974\times 10^{-13}J$

The energy released is $7.974\times 10^{-13}J$

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