Answer:
sorry but I can understand the question
Either one is fun and great to play!
<h2>
Answer:</h2>
The rate of deceleration is -0.14
<h2>
Explanation:</h2>
Using one of the equations of motion;
v = u + at
where;
v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)
u = initial velocity of the boat = 25m/s
a = acceleration of the boat
t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes
<em>Convert the time t = 3 minutes to seconds;</em>
=> 3 minutes = 3 x 60 seconds = 180seconds.
<em>Substitute the values of v, u, t into the equation above. We have;</em>
v = u + at
=> 0 = 25 + a(180)
=> 0 = 25 + 180a
<em>Make a the subject of the formula;</em>
=> 180a = 0 - 25
=> 180a = -25
=> a = -25/180
=> a = -0.14
The negative value of a shows that the boat is decelerating.
Therefore, the rate of deceleration of the speed boat is 0.14
Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation



![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99