Answer:

Explanation:
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In this case, we compute the heat output from coal, given its heating value and the mass flow:

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

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Answer:
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Explanation:
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Answer: true
Explanation:
it flows faster over the top of the wing because the top is more curved than the bottom of the wing. However
Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length
for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m
Answer:
D
Explanation:
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