Explanation & answer:
Given:
Fuel consumption, C = 22 L/h
Specific gravity = 0.8
output power, P = 55 kW
heating value, H = 44,000 kJ/kg
Solution:
Calculate energy intake
E = C*P*H
= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)
= (22/3600)*1000*0.8*44000 j/s
= 215111.1 j/s
Calculate output power
P = 55 kW
= 55000 j/s
Efficiency
= output / input
= P/E
=55000 / 215111.1
= 0.2557
= 25.6% to 1 decimal place.
What is the question asking?
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.
Answer:
The approximate magnitude of the force of air resistance is 540 N.
Explanation:
