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nevsk [136]
3 years ago
12

Which of the following is an example of benefits?

Physics
1 answer:
dangina [55]3 years ago
6 0
Insurance would be a example of benefits because you don’t receive insurance like you would salary, salary is a base thing you receive when you get a job while insurance you only get as a benefit with certain jobs
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Ignoring reflection at the air-water boundary, if the amplitude of a 10 GHz incident wave in air is 20 V/m at the water surface,
dimaraw [331]

Answer:

0.80267 m

Explanation:

E(z) = Electric field = 1 µV/m

E_0 = 20 V/m

z = Depth

\sigma = Conductivity = 0.1 S/m

\epsilon_r = 81

\mu = Impedance of free space = 120\pi\ \Omega

Frequency is given by

E(z)=E_0e^{-\alpha z}

Parameter is given by

\alpha=\dfrac{\sigma}{2}\sqrt{\dfrac{\mu}{\epsilon_r}}\\\Rightarrow \alpha=\dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{81}}\\\Rightarrow \alpha=20.94395\ N_p/m

From the first equation

1\times 10^{-6}=20e^{-20.94395z}\\\Rightarrow ln\dfrac{1\times 10^{-6}}{20}=-20.94395z\\\Rightarrow z=\dfrac{ln\dfrac{1\times 10^{-6}}{20}}{-20.94395}\\\Rightarrow z=0.80267\ m

The depth is 0.80267 m

7 0
3 years ago
when the c4 key on a piano keyboard is pressed, a string inside the piano is struck by a hammer and begins vibrating back and fo
Ugo [173]
The correct answer is f= 260 Hz
7 0
3 years ago
Read 2 more answers
What else is produced when sodium carbonate decomposes?
kompoz [17]

Answer: What else is produced when sodium carbonate decomposes?

Explanation:

3 0
2 years ago
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When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times l
nordsb [41]
<span>Depth = 5.0 Ă— 10^2 m
 Density of sea water = 1.025 x 10^3
 Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
  Since the normal pressure is retained in the hull, no need to bother about Po Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
 So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
 So it is 49.56 times larger.</span>
5 0
3 years ago
Read 2 more answers
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m
Reil [10]

Answer:

\Delta l=0.015m

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area A=0.5mm^2=0.5\times 10^{-6}m^2

Young's modulus \gamma=2\times 10^{11}Pa

Force F = 1500 N

So stress =\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa

We know that young's modulus =\frac{stress}{strain}

So 2\times 10^{11}=\frac{3\times 10^{9}}{strain}

strain=1.5\times 10^{-2}=0.015m

Now strain =\frac{\Delta l}{l}

0.015=\frac{\Delta l}{1}

\Delta l=0.015m

6 0
3 years ago
Read 2 more answers
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