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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 unusually bright centers found in some galaxies are called ______.

Nimfa-mama [501]

<span>Active Galactic Nuclei.</span>

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3 years ago

If you need 40.0 Nm of torque in order to loosen a nut on a wn

KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

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3 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$

Replacing,

$a =\frac{42N}{83000kg}$

$a =5.06*10^{-4}m/s^2$

The total speed change

$\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity }$ we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

$a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}$

We have that,

$t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }$

$t = 1403.16s$

Therefore the Rocket should be fired around to 1403.16s

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3 years ago

A Universe Question!

vfiekz [6]

I'm pretty sure this is A

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3 years ago

In a perfectly elastic collision between two perfectly rigid objects

ipn [44]

Both the total momentum and the total kinetic energy are conserved

Explanation:

- In a collision between two or more objects, if there are no external forces acting on the system (isolated system), the total momentum of the objects is always conserved. This is called principle of conservation of momentum, and can be written as follows:

$mu+MU = mv+MV$

where

m, M are the masses of the two objects

u, U are the initial velocities of the two objects

v, V are the final velocities of the two objects

- The total kinetic energy, however, is not always conserved. In fact, we have two types of collision:

1) In a perfectly elastic collision, the total kinetic energy of the objects is conserved. This means that we can write the following equation:

$\frac{1}{2}mu^2 + \frac{1}{2}MU^2 = \frac{1}{2}mv^2+\frac{1}{2}MV^2$

2) In an inelastic collision, the total kinetic energy of the object is NOT conserved. This means that part of the total kinetic energy is "lost", converted into other forms of energy (mainly thermal energy, due to the presence of frictional forces within the system). The most extreme case is called perfectly inelastic collision, in which the two objects stick together after the collision, and there is the maximum loss of kinetic energy.

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

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3 years ago