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Tju [1.3M]
3 years ago
5

What are the wavelength ranges for the following? (a) the AM radio band (540–1600 kHz) maximum wavelength m minimum wavelength m

(b) the the FM radio band (88–108 MHz) maximum wavelength m minimum wavelength m
Physics
1 answer:
Pie3 years ago
3 0

Answer:

Explanation:

a ) AM radio band (540–1600 kHz)

frequency = 540 kHz = 540 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 540 x 10³

= 555.55 m

frequency = 1600 kHz = 1600 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 1600 x 10³

= 187.5  m

maximum wavelength  =   555.55 m

minimum wavelength =  187.5 m

b )

AM radio band (88 - 108 MHz)

frequency = 88 MHz = 88 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 88 x 10⁶

= 3.41 m

frequency = 108 MHz = 108 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 108 x 10⁶

= 2.78  m

maximum wavelength  =   3.41 m

minimum wavelength =  2.78 m

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
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Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

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