Your answer is 311.29271 lbs
        
             
        
        
        
We have that the instantaneous velocity of the
shuttlecock when it hits the ground is

From the question we are told 
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration  is mathematically given as

Where
acceleration is still -9.81 m/s2,
Hence,

Therefore

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Answer:
0   (there is no speed)
Explanation:
If an object is at rest, it is not moving, and it doesn't have a speed, so the speed is zero.
 
        
                    
             
        
        
        
Answer:2.517 J/K
Explanation:
 Given
Reservoir 1 Temperature 
Reservoir 2 Temperature 
Let Q is the amount of heat Flows i.e. 
thus change in Entropy is given by 



                              
 
        
             
        
        
        
Answer:
The temperature of the steam during the heat rejection process is 42.5°C 
Explanation:
Given the data in the question;
the maximum temperature T
 in the cycle is twice the minimum absolute temperature T
 in the cycle 
T
  = 0.5T
 
now, we find the efficiency of the Carnot cycle engine
η
 = 1 - T
/T
η
 = 1 - T
/0.5T
η
 = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η
 = 1 - W
/Q
where W
 is net work done, Q
 is is the heat supplied
we substitute
0.5 = 60 / Q
Q
 = 60 / 0.5
Q
 = 120 kJ
Now, we apply the first law of thermodynamics to the system
W
 = Q
 - Q
60 = 120 -  Q
Q
 = 60 kJ
now, the amount of heat rejection per kg of steam is;
q
 = Q
/m
we substitute
q
 = 60/0.025
q
 = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q
 = h
 = 2400 kJ/kg
now, at  h
  = 2400 kJ/kg from saturated water tables;
T
 = 40 + ( 45 - 40 ) ( 
 )
T
 = 40 + (5) × (0.5)
T
 = 40 + 2.5
T
 = 42.5°C   
Therefore, The temperature of the steam during the heat rejection process is 42.5°C