Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 4.86 and angle θA = 23.5 ∘ measured in the sense from the +x
- axis toward the +y - axis, B⃗ with B = 4.02 and θB = 60.1 ∘, and C⃗ with magnitude C = 6.02 and in the +z - direction. Vectors A⃗ and B⃗ are in the xy-plane.
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Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity = 60 mph = 26.8224 m/s
Final velocity = 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = m(
² -
² )
we substitute
Δk = ×1814.37( (26.8224)² - (13.4112)² )
Δk = × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh