If the train stops then the math is gonna be c4
Answer:
435.032 kj
Explanation:
We can describe Heat transfer as a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.
Please refer to the attached file for the detailed step by step solution of the given problem
Answer:
give the question in English it'd be understandable
Answer:
![V_{p (load)} = 28,3 V - 0,7 V = 27,6 V](https://tex.z-dn.net/?f=V_%7Bp%20%28load%29%7D%20%3D%2028%2C3%20V%20-%200%2C7%20V%20%3D%2027%2C6%20V)
![V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V](https://tex.z-dn.net/?f=V_%7Bp%20%28load%29%7D%20%3D%2027%2C6%20V%5C%5CV_%7Bavg%7D%20%3D%2017%2C57%20V)
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
![I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA](https://tex.z-dn.net/?f=I_%7Bavg%7D%20%3D%20%5Cfrac%7BV_%7Bavg%7D%7D%7BR_%7Bload%7D%7D%20%3D%20%5Cfrac%7B17%2C57%20V%7D%7B1000%20%CE%A9%7D%20%3D%2017%2C57%20mA)
Explanation:
The peak voltage after the 6 to 1 step down is
. Then, the peak voltage of the rectified output is
V_{d}[/tex] and according to the statement, the diodes can be modeled to be
. Then, the peak voltage in the load is
.
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
The average output voltage is calculated as:
![V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V](https://tex.z-dn.net/?f=V_%7Bp%20%28load%29%7D%20%3D%2027%2C6%20V%5C%5CV_%7Bavg%7D%20%3D%2017%2C57%20V)
The average current in the load is calculated as:
![I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA](https://tex.z-dn.net/?f=I_%7Bavg%7D%20%3D%20%5Cfrac%7BV_%7Bavg%7D%7D%7BR_%7Bload%7D%7D%20%3D%20%5Cfrac%7B17%2C57%20V%7D%7B1000%20%CE%A9%7D%20%3D%2017%2C57%20mA)
Answer:it took 4 1/8th teaspoons
Explanation: