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3 years ago

11

If 20 beats are produce with one second which of the following frequency could possibly be held by two sound waves traveling thr

ough a medium a long a common path at the same time

1 answer:

musickatia [10]3 years ago

8 0

ANY two frequencies whose difference is 20 Hz.

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The anemometer is spinning and the pressure us high

skelet666 [1.2K]

Answer:

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2 years ago

Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive or attractive only

Y: Very small range

Z: Repulsive and attractive

8 0

3 years ago

Why are noise considerations important in optical fiber communications? 3. Describe the principle of "population inversion".

pochemuha

Answer and Explanation:

the electronic devices always have some noises present in the signal

there are some important considerations in optical fiber communications these are.

the noise which is contributed by transmitter are electronic random noise, low frequency noise
noise which is contributed by laser are relative intensity noise, mode partition noise, conversion of phase noise to amplitude noise.
noise contributed by photo detector are quantum shot noise, shot noise from dark current, avalanche multiplication noise.

PRINCIPLE OF POPULATION INVERSION :

The principle of population inversion is defined as for production of high percentage of simulated emission for a laser beam the number of atoms in higher state should be greater than lower energy state

7 0

3 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

2 years ago

g An electric hot plate raises its own internal energy and the internal energy of a cup of water by 9000J, and there is at the s

DochEvi [55]

Answer:

11700j

Explanation:

add the two because the plate has to maintain the temp.

2700+9000=11700

3 0

2 years ago