Round 1468.12277 to the nearest whole number
1 answer:
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To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).
By definition the efficiency of the heat engine is
Where,
Temperature at the room
Temperature of the soup
The work done is defined as,
Where represents the input heat and at the same time is defined as
Where,
Specific Heat
The change at the work would be defined then as
On the other hand we have that the heat lost by the soup is equal to
The ratio between both would be,
Replacing with our values we have,
Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.
Answer:
a) 15.77 m/sec2
b) 13.3 deg
Explanation:
we are given;
Flea force = F1=1.07×10⁻5 N j
Breeze force = F2 = 1.14× 10⁻6 N (-j
mass of flea =6.0 ×10⁻7 kg
So net force on the flea=F1+F2+weight of flea=1.07×10⁻5 j +1.14× 10⁻6 i + 6.0 ×10⁻7 (-j) ×9.8= ma
==> ma = 1.07×10⁻5 j - 0.588×10⁻5 j + 0.114×10⁻5 i
==> ma= 0.482 ×10⁻5 j +0.114×10⁻5 i
==> ma = 0.114×10⁻5 i +0.482 ×10⁻5 j
== a = (0.114×10⁻5 i +0.482 ×10⁻5 j) / 6.0 ×10⁻7
==> a =
==>a= (1.9 j+8.03 i ) m/sec2
mag of a = 15.77 m/sec2
direction angle = tan⁻1(1.9/8.03)=13.3°