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1 year ago

A radio station broadcast on a frequency of 3.7 mhz what is the energy of the radio wave

1 answer:

3 0

A radio station broadcast on a frequency of 3.7 mhz what is the energy of the radio wave A radio station broadcasts its programmes at a wavelength of 500 m. Find the frequency of the radiowaves transmitted by the radio station, if the speed of radiowaves in air is 3 x 108 m/s. Ans: 6 x 10 Hz

<h3>What is radio station ?</h3>

Radio broadcasting is the act of sending audio (sound), occasionally together with accompanying metadata, across radio waves to radio receivers used by the general public. Unlike satellite radio, which uses a satellite in Earth's orbit, terrestrial radio broadcasting uses a land-based radio station to transmit radio waves. The listener needs a broadcast radio receiver to hear the material (radio). A radio network with which stations frequently have affiliations provide content in a standard radio format, whether through broadcast syndication, simulcasting, or both. Radio stations use a variety of modulations to transmit their signals, including FM (frequency modulation), which is an older analog audio standard, and AM (amplitude modulation).

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Elina [12.6K]

Answer:

The vertical distance that the ski jumper fell is 417.45 m.

Explanation:

Given;

initial horizontal velocity of the jumper, $V_x$ = 26 m/s

horizontal distance of the jumper, dx = 240 m

The time of the motion is given by;

dx = Vₓt

t = dx / Vₓ

t = 240 / 26

t = 9.23 s

The vertical distance traveled by the diver is given by;

$d_y = V_yt + \frac{1}{2}gt^2$

initial vertical velocity, $V_y$ , = 0

$d_y = \frac{1}{2}gt^2\\\\d_y = \frac{1}{2}(9.8)(9.23)^2\\\\d_y = 417.45 \ m$

Therefore, the vertical distance that the ski jumper fell is 417.45 m.

6 0

2 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

2 years ago

the brakes on a car do 240000j of work in stopping the car. if the car travels a distance of 40m while the brakes re being appli

Tpy6a [65]

A joule is one Newton of force applied over a meter.
For every meter, the brakes put 240000N of force (N=Newtons).
For 40m, multiply the Newtons by 40.
240000N*40=9600000N

3 0

2 years ago

Converting 15 miles to kilometer

Dmitry [639]

15 miles to kilometers would be: 24.14 kilometers

8 0

2 years ago

Read 2 more answers

Jeffery gains super strength and pushes two different objects with the same amount of force. Object A accelerates at 40 m/s2, an

Montano1993 [528]

Answer: Object B

Explanation: Acceleration is directly proportional to force and inversely proportional to mass. It implies that more massive objects accelerates at a slower rate.

5 0

3 years ago