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makvit [3.9K]
3 years ago
13

A TV satellite broadcasts at a frequency of 5000 MHz, (1 MHz = 1 million Hertz). What is the wavelength of this radiation?

Physics
2 answers:
Nitella [24]3 years ago
5 0

Answer:

\lambda=0.06\ m

Explanation:

Given:

  • frequency of the broadcast, f=5000\ MHz=5\times 10^9\ Hz
  • we have the speed of the radiation equal to the speed of light, c=3\times 10^8\ m.s^{-1}

The broadcast waves are the electromagnetic waves but it can travel only upto a hundred kilometers without any loss of information carried by it.

<u>The relation between the frequency and the wavelength:</u>

\lambda=\frac{c}{f}

\lambda=\frac{3\times 10^8}{5\times 10^9}

\lambda=0.06\ m

goldfiish [28.3K]3 years ago
5 0

Answer:

The wavelength of this radiation is 0.06 m.

Explanation:

Given that,

Frequency = 5000 MHz

We know that,

Speed of light c= 3\times10^{8}\ m/s

We need to calculate the wavelength of this radiation

Using formula of wavelength

\lambda=\dfrac{c}{\nu}

Where, \lambda = wavelength

\nu = frequency

c = speed of light

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{5\times10^{9}}

\lambda=0.06\ m

Hence, The wavelength of this radiation is 0.06 m.

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A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu
galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

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