Question 13 (1 point)
1 answer:
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To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
The extension of the spring due to the weight of the object on Moon is a value of , then
Recall that gravity on the moon is a sixth of Earth's gravity.
We have that the displacement at the earth was , then
Therefore the displacement of the mass on the spring on Moon is 0.05m
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