Answer:
(a) the subtransient current through the breaker in per-unit and in kA rms = 71316.39kA
(b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. = 152KA
Explanation:
check the attached files for explanation
This is a new technology meant to reduce vehicle roll-overs A. Lane-departure warning B. Anti-lock brakes C. Inspection reports
1 answer:
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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 with Explanation:
Part a)
The volume of water in the tank as a function of time is plotted in the below attached figure.
The vertical intercept of the graph is 46.
Part b)
The vertical intercept represents the volume of water that is initially present in the tank before draining begins.
Part c)
To find the time required to completely drain the tank we calculate the volume of the water in the tank to zero.
Part d)
The horizontal intercept represents the time it takes to empty the tank which as calculated above is 13.143 minutes.
