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Katarina [22]
2 years ago
14

To test the strength of a retainment wall designed to protect a nuclear reactor,

Physics
1 answer:
Hatshy [7]2 years ago
6 0

Answer:

The speed of the F-4 Phantom upon impact is approximately 214.92 m/s

Explanation:

The given parameter of the motion of the F-4 Phantom jet aircraft are;

The mass of the F-4 phantom rocket, m₁ = 19,100 kg

The mass of the retaining wall, m₂ = 469,000 kg

The velocity of the combined mass of the wall and the F-4 after collision, v₃ = 8.41 m/s

The retainment wall was initially at rest, therefore, v₂ = 0 m/s

Let 'v₁', represent the speed of the F-4 Phantom upon impact

By the principle of conservation of linear momentum, we have;

m₁·v₁ + m₁·v₂ = m₁·v₃ + m₂·v₃ = (m₁ + m₂)·v₃

Plugging in the values, we have;

19,100 × v₁ + 469,000 × 0  = (19,100 + 469,000) × 8.41

∴ v₁ = (19,100 + 469,000) × 8.41/(19,100) = 214.917329843

v₁ ≈ 214.92

The speed of the F-4 Phantom upon impact, v₁ ≈ 214.92 m/s.

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a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

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initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

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final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

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