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3 years ago

Identify a source of microwaves

Physics

2 answers:

siniylev [52]3 years ago

4 0

Answer:

Stars, including the Sun, are natural microwave sources.

Explanation: Under the right conditions, atoms and molecules can emit microwaves. Man-made sources of microwaves include microwave ovens, masers, circuits, communication transmission towers, and radar. Either solid state devices or special vacuum tubes may be used to produce microwaves.

brilliants [131]3 years ago

3 0

Explanation:

Microwave sources include artificial devices such as circuits, transmission towers, radar, masers, and microwave ovens, as well as natural sources such as the Sun and the Cosmic Microwave Background. Microwaves can also be produced by atoms and molecules.

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4. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the

vlada-n [284]

The net force on the sled is 300 N

Explanation:

First of all, we start by finding the acceleration of the bobsled, by using the suvat equation:

$v^2-u^2=2as$

where:

v = 6.0 m/s is the final velocity of the sled

u = 0 is the initial velocity

a is the acceleration

s = 4.5 m is the displacement of the sled

Solving for a, we find

$a=\frac{v^2-u^2}{2s}=\frac{6.0^2-0}{2(4.5)}=4 m/s^2$

Now we can find the net force on the sled by using Newton's second law:

F = ma

where

F is the net force

m = 75 kg is the mass of the sled

$a=4 m/s^2$ is the acceleration

Solving the equation, we find the net force:

$F=(75)(4)=300 N$

Learn more about acceleration and Newton laws here:

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8 0

3 years ago

A small steel wire of diameter 1.0 mm is connected to an oscillator and is under a tension of 7.5 N. The frequency of the oscill

avanturin [10]

Answer:

The output power is $P= 0.764Watt$

The Amplitude would decrease by $\frac{1}{2}$

Explanation:

From the question we are told that

The diameter of the steel wire is = 1.0mm $= \frac{1}{1000} = 1.0*10^{-3}m$

The raduis of this steel wire is $r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire this is mathematically represented as

$P = \frac{1}{2} \mu w^2 A^2 v ----(1)$

Where $\mu$ is the mass per unit length of the wire

This is mathematically evaluated as

$\mu = a* \rho$

Where a is the area of the the wire = $\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2$

$\rho$ is the density of steel with a generally value of $7850 kg/m^3$

$\mu = 7.855*10^{-7} *7850$

$= 6.162*10^{-3}kg/m$

$w$ is velocity of the wave

This is mathematically evaluated as

$w=2 \pi f$

substituting 60Hz for f

We have

$w = 2 *3.142 * 60$

$=377.04 \ rad/s$

$A$ is the amplitude with a given value of 0.50 cm $= \frac{0.50}{100} = 0.50 *10^{-2}m$

v is the linear velocity of the wave

This is mathematically evaluated as

$v = \sqrt{\frac{T}{\mu} }$

Where T is the tension with a given value of $7.5N$

$v = \sqrt{\frac{7.5}{6.162*10^{-3}} }$

$=34.89 m/s$

Substituting values into equation 1

$P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89$

$P= 0.764Watt$

Since the doubling of the frequency does not affect the amplitude and from equation one the output power is $\ \frac{1}{2}$ of the Amplitude, Then the Amplitude would decrease by $\frac{1}{2}$