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pochemuha
3 years ago
9

A bungee jumper falls 97.2 feet before bouncing back up at the end of his bungee cord. Then he falls 64.8 feet before bouncing u

p again. On his third descent, he falls 43.2 feet before going back up. What is the total distance the bungee jumper drops after five falls if the distances continue in this pattern? (Do not include the distance traveled up.) A)234.0 ft. B)253.2 ft. C)789.8 ft. D)1,281.8 ft
Physics
1 answer:
viktelen [127]3 years ago
3 0
The answer is letter B. 

The solution is as follows:
97.2 ft/ x = 64.8 ft
97.2 / 64.8 = x
x = 1.5

Divide 97.2 by x 4 times and we have the following data:
1st fall = 97.2 ft
2nd fall = 64.8 ft
3rd fall =  43.2 ft
4th fall =  28.8 ft
5th fall =  19.2 ft
Total distance of falls = 253.2 ft
 


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This is a computer program, why is there an error between the computer values and the nominal values of R?
ss7ja [257]

Answer:

the difference is due to resistance tolerance

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   This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values ​​and the values ​​measured with the instruments, multimeters.

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In summary, the difference is due to resistance tolerance.

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4 years ago
What happens to the resistance of a wire as it gets wet
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8 0
3 years ago
Potassium is a crucial element for the healthy operation of the human
Degger [83]

Answer:

1

  The mass of the Potassium-40 is  m_{40}} = 2.88*10^{-6} kg

2

  The Dose per year in Sieverts is   Dose_s = 26.4 *10^{-10}

Explanation:

From the question we are told that

   The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

    Their abundance is 93.26%, 0.012% and 6.728%

   The mass of potassium contained in human body is  m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg per kg of the body

    The mass of the first body is  m_1 = 80 \ kg

Now the mass of  potassium  in this body is mathematically evaluated as

       m_p =  m * m_1

substituting value

       m_p =  80  * 0.0003

      m_p  =0.024 kg

The amount of Potassium-40 present  is mathematically evaluated as

      m_{40}} =0.012% * 0.024

      m_{40}} = \frac{0.012}{100}  * 0.024

      m_{40}} = 2.88*10^{-6} kg

The dose of energy absorbed per year is mathematically represented as

          Dose  = \frac{E}{m_1}

Where E is the energy absorbed which is given as E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}

    Substituting value

            Dose  = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}

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The Dose in Sieverts is evaluated as

       Dose_s = REB * Dose

       Dose_s = 1.2 * 22*10^{-10}

       Dose_s = 26.4 *10^{-10}

             

3 0
3 years ago
An AC circuit has an RMS voltage of 80 VAC. What's the circuit's average voltage?
yaroslaw [1]

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So the peak value is (√2) x (RMS value)

-- The Average value of an AC waveform is (2/π) x (peak value)

So the peak value is (π/2) x (Average value).

-- So far, this is all very entertaining, but how does it help us answer the question.

Well, we found the peak value in terms of the RMS and in terms of the Average.  So we can set these equal to each other, and solve for the Average in terms of the RMS.  This sounds like such a good plan, I think I'll do it !

Peak = (√2) x (RMS value)  and  Peak also = (π/2) x (Average value).

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Divide each side by π :  (√2) · RMS / π = (1/2) · Average

Multiply each side by 2 :  Average = (2/π) · (√2) · RMS .

You said that the RMS value is 80 V, so

Average = (2/π) · (√2) · (80)

Average = (2 · √2 / π) · (80)

Average = 160√2 / π

<em>Average = 72 volts </em>.  (But be sure to read the 'gotcha' below.)


Now I'll go ahead and tell you the 'gotcha':

All of these numbers are true, as far as they go.  But the 'average' is only true for 1/2 cycle of an AC wave.  Picture an AC wave in your mind.  You'll see that it spends just as much time being negative as it spends being positive.  So the 'average' of any number of AC <em><u>whole cycles</u></em> is <em>zero.</em>

7 0
3 years ago
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