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2 years 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).

To learn more about radio station from the given link:

brainly.com/question/26439029

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A typical mattress has a network of springs that provide support. If you sit on a mattress, the springs compress. A heavier pers

GenaCL600 [577]

Answer:

Explanation:

Spring has a tendency to store energy in them and deform its shape when force is applied on it. Once the applied force is removed it regains its original shape and size.

It is in helical shape and is used in mattress to give structure and support. Spring have elastic nature and follows spring forces, F = k * x

where is the applied force, k is the spring constant and x is the amount of extension.

When a heavier person sits on a mattress, more weight is applied on springs and they form coils, as weight is removed they regains its shape again.

4 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?

kumpel [21]

The SI unit for heat energy is joule

4 0

3 years ago

What is the mass of 5 moles of gold

Serhud [2]

First of all, we know that one mole is equal to the atomic number of an element.

The atomic number of gold is 197.0g Au

And we need to find 5 moles.

5 * 197.0 g Au = 985.0g

Grams is used to measure mass.

Answer: 985.0g

8 0

3 years ago

Read 2 more answers

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

Leona [35]

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

$wsin\theta-\mu F_n=ma$ That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

6 0

3 years ago