The correct options are:
D
"Radio waves have a lower frequency, which makes them safer for humans."
B
"Radio waves take less energy to produce."
<h3>
Why do we radio waves over other electromagnetic waves to transmit information to Earth? </h3>
Radio waves are electromagnetic waves with frequencies on the range from 10 KHz to 10 THz.
Now, remember that all electromagnetic waves have the same speed, which is the speed of light, and the energy of a wave is proportional to its frequency.
Particularly, we can see that radio waves have small frequencies (smaller than infrared light) so these waves carry very little energy.
With that in mind, the correct options are.
D
"Radio waves have a lower frequency, which makes them safer for humans."
B
"Radio waves take less energy to produce."
These are the two main reasons of why we use radio waves.
If you want to learn more about electromagnetic waves.
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Answer:
The maximum amount of work is
Explanation:
From the question we are told that
The temperature of the environment is
The volume of container A is
Initially the number of moles is
The volume of container B is
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
Now from the Ideal gas law
So substituting for in the equation above
Where R is the gas constant with a values of
Substituting values we have that
Answer:
95 J
Explanation:
You can calculate efficiency by dividing useful output by total input, then multiplying it to 100.
So the foumula goes like:
Efficiency= (Useful output/Total input)x100
In this question,
Efficiency= 95%
Useful output= x
Total input= 200
Therefore;
95=(x/200)x100
0.95=x/100
x=0.95x100
x=95 Joules
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
The answer is bend towards normal.