Technician A says that coolant freeze protection can be measured with a
2 answers:
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<u>Explanation:</u>
5 Horsepower for 30 mins,
(5)(745.7) = 3.7285 KW power delivered
General Efficiency of IC engine = 20%
Power required =
Energy required per week,
=P × Time = 18.64 × 60 × 30 = 33.5565 MJ
Lawn area = (30) (20) = 600
let sunlight hours be 8 hours
Hence, solar power input on lawn,
=5.62×3600 = 20232 kJ//day
energy input in lawn = (600) (20232) (7)
= 84974.4 mJ/week
Chemical efficiency by photosynthesis = 4%
Chemical content in grass = (84974.4) (0.04)
= 3398.97 mJ
Mass of the clippers
pounds
Removing water content,
dried grass clippings
= 11533.25 gallons
Trash cans repaired
By burning the gas, total energy input = 3398.97 MJ × 0.2
= 679.794 MJ
Efficiency of steeling engine = 20%
Energy output by engine = 679.794 ×0.2
= 135.96 mJ
Energy required by mover = 33.5565 mJ
Hence, Energy (output) ⇒ energy required
Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
=
By putting values, we get
=
=
=
hence,
The average force will be:
=
=
=
Or,
=