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ira [324]
2 years ago
14

Imagine a Carnot engine has a hot reservoir of 680 K and a cold reservoir of 220 K. What is the efficiency of the engine? 58.3%

75.9% 62.3% 72.1% 67.6%
Physics
2 answers:
Strike441 [17]2 years ago
7 0
N = 1 - (QC/QH)

n = efficiency

QC = Cold Reservoir = 220 K
QH = Hot Reservoir = 680 K

n = 1 - (220 \div 680) (\div 10)\\ n = 1 - (22 \div 68) (\div 2) \\ n = 1 - (11 \div 34) \\ n = (34 - 11) \div 34 \\ n = 23 \div 34 \\ n = 0.676... \times 100\\ n = 67.6

Answer: The engine's efficiency is 67.6% .
Romashka [77]2 years ago
6 0

the other person's right, it's E. 67.6% for plato users :)

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Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
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Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

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Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

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\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

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