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CaHeK987 [17]
3 years ago
9

What do radio waves and microwaves have in common?

Physics
2 answers:
Lunna [17]3 years ago
8 0

Both are at the side of the spectrum that has the lower frequency

Free_Kalibri [48]3 years ago
6 0

Answer: Both are at the side of the spectrum that has the lowest frequency.

Explanation:-

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They travel with the speed of light.

Both radio and microwaves are electromagnetic waves which are adjacent to each other in electromagnetic spectrum towards lower frequency and higher wavelength.

The relationship between wavelength and frequency of the wave follows the equation:

\nu=\frac{c}{\lambda}

where,

\nu = frequency of the wave

c = speed of light  

\lambda = wavelength of the wave

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OC, The Tariff of Abominations caused conflicts between the North and the South regarding their sectional interests.

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The Tariff of Abominations did cause conflicts because it widened income inequality since it favored the rich upper class while burdening the lower class. The South was the lower class, and the North was more of the upper class.

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What happened on may 24 2002 in history
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Russia and the United States sign the Moscow Treaty.
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3 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
If you were given distance and period of time, what can you calculate?
antiseptic1488 [7]

If you were given distance & period of time, you would be able to calculate the speed.

Hope this helps!

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