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makvit [3.9K]
2 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]2 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]2 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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Law Incorporation [45]

Answer:

5.32\cdot 10^4 g

Explanation:

First of all, we need to find the volume of the room, which is given by

V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3

Now we  can find the mass of the air by using

m=dV

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d=1.29 g/dm^3 is the density of the air

V=41.3 m^3 = 41,300 dm^3 is the volume of the room

Substituting,

m=(1.29)(41300)=5.32\cdot 10^4 g

6 0
2 years ago
A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m
Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

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Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

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Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

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A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath
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Answer:

A

Explanation:

Solution:-

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- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

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On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the
EastWind [94]

Answer:

ω = √(2T / (mL))

Explanation:

(a) Draw a free body diagram of the mass.  There are two tension forces, one pulling down and left, the other pulling down and right.

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For small angles, sin θ ≈ tan θ.

∑F = -2T tan θ

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(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m).  (This is derived by solving a second order differential equation.)

In this case, k = 2T/L, so the frequency is:

ω = √((2T/L) / m)

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