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3 years ago

What is the practical applications of radio waves

1 answer:

5 0

Radio waves have many uses—the category is divided into many subcategories, including microwaves and electromagnetic waves used for AM and FM radio, cellular telephones and TV.

The lowest commonly encountered radio frequencies are produced by high-voltage AC power transmission lines at frequencies of 50 or 60 Hz. These extremely long wavelength electromagnetic waves (about 6000 km) are one means of energy loss in long-distance power transmission.

Extremely low frequency (ELF) radio waves of about 1 kHz are used to communicate with submerged submarines. The ability of radio waves to penetrate salt water is related to their wavelength (much like ultrasound penetrating tissue)—the longer the wavelength, the farther they penetrate. Since salt water is a good conductor, radio waves are strongly absorbed by it; very long wavelengths are needed to reach a submarine under the surface.

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The name of the SI unit force

Lelechka [254]

Answer: Newton

Explanation:-

Force is defined as an interaction that causes an affected object to be pushed or pulled in a certain direction.

$Force=m\times a$

where m= mass of an object

a = acceleration of the object

S.I unit of mass is kg and S.I units of acceleration is $ms^{-2}$

Thus For S,I units of force:

$Force=kg\times ms^{-2}=1Newton$

3 0

4 years ago

Read 2 more answers

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

$v = \frac{{2\pi L}}{T}$

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

$v = \frac{{2\pi L}}{T}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{2\pi }}{T}$ is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute $v = \frac{{2\pi L}}{T}$ in the force expression ${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$ as follows:

$\begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{4{\pi ^2}}}{{{T^2}}}$ not affect the force value for six satellites.Therefore, we can write the expression of ${F_c}$ given as follows:

${F_c} = kmL$

Here, k refers to constant value and equal to $\frac{{4{\pi ^2}}}{{{T^2}}}$

${F_A} = k{m_A}{L_A}$

Substitute 200 kg for ${m_A}$ and 5000 m for LA in the expression ${F_A} = k{m_A}{L_A}$

$\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite B from their rocket is given as follows: ${F_B} = k{m_B}{L_B}$

Substitute 400 kg for ${m_B}$ and 2500 m for in the expression ${F_B} = k{m_B}{L_B}$

$\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite C from their rocket is given as follows: ${F_C} = k{m_C}{L_C}$

Substitute 100 kg for ${m_C}$ and 2500 m for in the above expression ${F_C} = k{m_C}{L_C}$

$\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite D from their rocket is given as follows: ${F_D} = k{m_D}{L_D}$

Substitute 100 kg for ${m_D}$ and 10000 m for ${L_D}$ in the expression ${F_D} = k{m_D}{L_D}$

$\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite E from their rocket is given as follows: ${F_E} = k{m_E}{L_E}$

Substitute 800 kg for ${m_E}$ and 5000 m for ${L_E}$ in the expression ${F_E} = k{m_E}{L_E}$

$\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite F from their rocket is given as follows: ${F_F} = k{m_F}{L_F}$

Substitute 300 kg for ${m_F}$ and 7500 m for ${L_F}$ in the expression ${F_F} = k{m_F}{L_F}$

$\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The value of forces obtained for the six-different satellite are as follows.

$\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

7 0

4 years ago

A farmer uses 3450 J of energy on a wheel and axle to lift a 15 kg bucket of water 20 m to the top of her well. What is the effi

Anika [276]

<span>Efficiency is the measure of how efficient a process is. It is used to assess the ability of a process in avoiding waste energy, materials, money and time in doing a desirable output. It is calculated as;

Efficiency = useful energy ouput / total energy input</span>

<span>Efficiency = 15(9.81)(20) / 3450 = 0.85</span>

8 0

3 years ago

Read 2 more answers

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t

Lelu [443]

Answer:

c) At a distance greater than r

Explanation:

If G= Gravitational constant

M= Mass of earth

r= distance from earth center

then orbital speed is ;

v = $\sqrt{\frac{GM}{r} }$

==> v²=GM/r

If speed of first satellite = V₁

==> V₁² = GM/r

==> r = GM/V₁²

If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

So our answer is c

5 0

3 years ago

What was the trail of tears?

Deffense [45]

DescriptionThe Trail of Tears was a series of forced relocations of approximately 60,000 Native Americans in the United States from their ancestral homelands in the Southeastern United States, to areas to the west of the Mississippi River that had been designated as Indian Territory.

4 0

3 years ago