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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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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
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Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
1 year ago
A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction
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Answer:

F=UR

52.9=U*46.6

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U=1.135

4 0
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Answer: 2R

Explanation:

Here the person travels пR distance. We know that the circumference of a circle is 2πR. So your imaginated person has traveled the distance which​ is half of the circumference of the circle. And this distance is equal to its diameter. We know that diameter of a circle is two times larger than the radius. So the person's displacement is two times of the radius, means 2R. [Here 'R' means the radius of the circle]

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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
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I attached the missing picture.
We can figure this one out using the law of conservation of energy.
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A: mgh_1+\frac{mv_1^2}{2}
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
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So, we know that the car is approaching the point B with the following amount of energy:
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B: mgh_2+\frac{mv_2}{2}
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mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
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Which of the following characteristics of Earth's relationship to the Sun explains the existence of Earth's seasons? Choose all
Aneli [31]

Answer:

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Explanation:

The earth tilt at an angle causes the sun rays to hit the earth surface around the globe differently. Due to the oblique angle that the rays hit the subtropics and poles, there is less heat intensity compared to the equator where the sun rays hit the earth's surface at a more or less right angle.

The earth rotation around the sun also causes seasons coupled with the earth’s tilts. As the earth rotates, in one point in the orbit, the northern or southern hemispheres will be tilted towards the sun. The phenomenon varies the local temperatures of particular regions of the earth hence driving seasonal climatic changes.

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