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Vikentia [17]
3 years ago
11

An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?

Engineering
1 answer:
kifflom [539]3 years ago
7 0

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = \frac{NE}{t}

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = \frac{NE}{t} = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}          [t = 1 s for per second]

or

550 × 10³ = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

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Write a method called letterCount that takes two String arguments, one containing some text and the other containing a single le
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Answer:

I am writing a Python program.  Here is the function letterCount. This function takes two string arguments text and letter and return count of all occurrences of a letter in the text.

def letterCount(text, letter):  

 count = 0  # to count occurrences of letter in the text string

 for char in text:  # loop moves through each character in the text

   if letter == char: # if given letter matches with the value in char

     count += 1  # keeps counting occurrence of a letter in text

 return count # returns how many times a letter occurred in text

   

Explanation:

In order to see if this function works you can check by calling this function and passing a text and a letter as following:

print(letterCount('apples are tasty','a'))

Output:

3

Now lets see how this function works using the above text and letter values.

text = apples are tasty

letter = a

So the function has to compute the occurrences of 'a' in the given text 'apples are tasty'.

The loop has a variable char that moves through each character given in the text (from a of apples to y of tasty) so it is used as an index variable.

char checks each character of the text string for the occurrence of letter a.

The if condition checks if the char is positioned at a character which matches the given letter i.e. a. If it is true e.g if char is at character a of apple so the if condition evaluates to true.

When the if condition evaluates to true this means one occurrence is found and this count variable counts this occurrence. So count increments every time the occurrence of letter a is found in apples are tasty text.

The loop breaks when every character in text is traversed and finally the count variable returns all of the occurrences of that letter (a) in the given text (apples are tasty). As a occurs 3 times in text so 3 is returned in output.

The screen shot of program along with output is attached.

4 0
3 years ago
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Answer:

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3 years ago
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Answer:

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Explanation:

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Answer:

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2 years ago
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A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov
babunello [35]

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H_{pump = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D  = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H_p = (p/ρq)₂ + (v²/2g)₂ + z₂ + h_L

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[  ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH_pQ / η

where; H_p = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = (P_{atom / ρg) - h_s - ( P_v / ρg )

we substitute

25  = ( 2116 / 62.42 ) - h_s - ( 30 / 62.42 )

h_s = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

6 0
3 years ago
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