Answer:
Explanation:
a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)
work done raised the potential energy
b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W
c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%
Not a very likely scenario.
Letter D: The amount of sleep a student gets affects student achievement
Answer: A <u>Nebula </u>is left behind. A spectacular explosion in which a star ejects most of its mass in a violently expanding cloud of debris.
Hope this helps!
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>
Magna Carta. I’m pretty sure it’s the correct answer.
