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Evgen [1.6K]
3 years ago
14

Suppose a square wave signal has a 65 percent duty cycle and an on-state voltage of 40 volts DC. What is the average DC voltage

Physics
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

The voltage is \= DC _v  =    2.6 \  V

Explanation:

From the question we are told that

  The duty cycle is  p =  65% = 0.65

   The on - state voltage is  V  =  40 \  volt

   

Generally the average DC voltage is mathematically represented as

        \= DC _v  =   p *  V

=>      \= DC _v  =     40 * 0.65

=>      \= DC _v  =    2.6 \  V

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Atoms are known to form charges due to their ability to gain and lose electrons in reactions. As a result, atoms
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Answer:

C. Fe: 24 electrons

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   O: 10 electrons: 8 protons = 8 - 10 = -2

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List five energy sources that are used in the united states that prduce electrical energy
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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
Sever21 [200]

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

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