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hram777 [196]
3 years ago
11

The speed of an airplane doubles in flight. (a) how is the law of conservation of momentum obeyed in this situation? (b) the law

of conservation of energy?
Physics
1 answer:
telo118 [61]3 years ago
7 0
<span>a) The law of conservation of momentum is satisfied because there must be air (and other combustion byproducts) expelled behind the aircraft as it moves forward. The collective momentum of those gases is equal in magnitude and opposite in direction to the forward momentum of the aircraft. b) The aircraft obeys conservation of energy because it is simply converting energy to different forms. In its stationary form, there existed latent chemical energy in both the fuel on board and the oxygen in the air. As the aircraft moves, some of that chemical energy is converted into kinetic energy and heat, but it is indeed conserved.</span>
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As we know that

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

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\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}

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A gold bar 20.0kg at 35.0°c is placed in a large insulated 0.8kg glass container at 15°c and 2.0kg of water at 25°c.. calculate
Oksanka [162]

Answer:

The final equilibrium temperature is approximately 26.69 °C

Explanation:

The heat transferred, ΔQ, from a hot body to a cold one is given by the following formula;

ΔQ = m·c·ΔT

Where;

m = The mass of the body

c = The specific heat capacity of the body

ΔT = The temperature change of the body

The given mass of the gold bar, m₁ = 20.0 kg

The initial temperature of the gold bar, T₁ = 35.0 °C

The specific heat capacity of gold, c₁ = 0.13 kJ/(kg·K)

The mass of the glass container, m₂ = 0.8 kg

The initial temperature of the glass container, T₂ = 15°C

The specific heat capacity of glass, c₂ = 0.792 kJ/(kg·K)

The mass of the added water, m₃ = 2.0 kg

The initial temperature of the added water, T₃ = 25°C

The specific heat capacity of water, c₃ = 4.2 kJ/(kg·K)

The heat lost by the gold = The heat gained by the glass and the water

Let 'T' represent the temperature at the final equilibrium, we have;

m₁·c₁·ΔT₁ = m₂·c₂·ΔT₂ + m₃·c₃·ΔT₃

Where;

ΔT₁ = T₁ - T

ΔT₂ = T - T₂

ΔT₃ = T - T₃

∴ 20.0 × 0.13 × (35 - T) = 0.8 × 0.792 × (T - 15) + 2.0 × 4.2 × (T - 25)

Expanding and collecting like terms (using a graphing calculator) gives;

91 - 2.6·T = 9.0336·T - 219.504

9.0336·T + 2.6·T = 219.504 + 91 = 310.504

11.6336·T = 310.504

T = 310.504/11.6336 ≈ 26.69

The final equilibrium temperature, T ≈ 26.69 °C.

4 0
3 years ago
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