Answer:
(a): The launch velocity is Vx= 666.66 m/s.
(b): The angle wich the projectile contacts the ground is α= 0.38°
Explanation:
h= 1m
g= 9.8 m/s²
h= g*t²/2
t= 0.45 s
Vy= g*t
Vy= 4.42 m/s
d=Vx* t
Vx= 666.66 m/s (a)
α= tg⁻¹ ( Vy/Vx)
α= 0.38° (b)
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%

W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000

<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
I believe number 4 I could be wrong but I think it’s 4