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Rainbow [258]
3 years ago
7

heat engine operating between 100 °C and 700 °C has an efficiency equal to 40% of the maximum theoretical efficiency. How much e

nergy does this engine extract from the hot reservoir in order to do 5000 J of mechanical work?
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

Explanation:

Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.

= 600 / (273 + 700 )

= 600 / 973

= .6166

operating efficiency = 40% of .6166

= .4 x .6166

= .2466 = 24.66 %

efficiency = work output / heat input

= 5000 / heat input = .2466

heat input = 5000 / .2466

= 20275.75 J .

HEAT EXTRACED = 20275.75 J.

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A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

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Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
3 years ago
How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?
Hitman42 [59]

Answer:

C. The block and the sphere would have the same weight.

Explanation:

If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.

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3 years ago
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Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might
Tanya [424]
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3 years ago
How do we use energy transformation in our daily lives?
olga55 [171]

Answer:hat are some examples of energy transformation?

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Explanation:

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a current of 5 ampere is passed for 2 hours in an electric iron having a resistance of 100 ohms calculate the heat produced
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Answer:

\boxed{\sf Heat \ produced \ (H) = 5 \ kWh}

Given:

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Current (I) = 5 A

Time (t) = 2 hours

To Find:

Heat developed (H) in the electric iron

Explanation:

Formula:

\boxed{ \bold{ \sf H = I^2Rt}}

Substituting values of I, R & t in the equation:

\sf \implies H =  {5}^{2}  \times 100 \times 2 \\  \\ \sf \implies H = 25 \times 100 \times 2 \\  \\  \sf \implies H = 5000  \\  \\ \sf \implies H = 5 \: kWh

\therefore

Heat developed (H) in the electric iron = 15 kWh

3 0
3 years ago
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