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.
Horizontal distance covered by a projectile is X = Vix *T
where Vix is the initial horizontal component of velocity and T is time taken by the projectile
Vix = ViCos theta
In question they said that initial velocity and angle is same on earth and moon
so Vix would remains same
now let's see about time taken T
time taken to reach the highest point
Vfy = Viy +gt
at highest point vertical velocity become zero so Vfy =0
0 = Vi Sin theta + gt
t = Vi Sintheta /g
Total time taken to land will be twice of that
On earth
Te= 2t
Te = 2Sinθ/g
on moon g is one-sixth of g(earth)
Tm = 2Sinθ/(g/6)
Tm = 6(2Sinθ/g)
Tm = 6Te
so total time taken by the projectile on moon will be six times the time taken on earth
From first equation X = Vix*T
we can see that X will also be 6 times on moon than earth
so projectile will cover 6 times distance on moon than on earth
Define displacement and I'll help you