At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
a) [tex]P = 75 \times (2.7\times10^7) = 2.025\times10^9 W[/tex]
b) [tex]Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%[/tex]
b) [tex]Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%[/tex]
Answer:
jj
Explanation:
Energy obtained by burning of 1 kg coal = 27 million Joule
a.
Energy obtained by burning of 75 kg coal = 27 x 75 = 2025 million Joule
Power = Energy obtained per second
Power = [tex]\frac{2025}{1} million Joule per second[/tex]
P = [tex]\frac{2025}{1}\times 10^{6} Watt[/tex]
P = [tex]\frac{2.025}{1}\times 10^{9} Watt[/tex]
b.
Input power = 2025 million Watt
Output power = 800 million Watt
Efficiency = output power / input power
Efficiency = 800 / 2025 = 0.395
Efficiency = 39.5%
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.