Answer:
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg 
Explanation:
Total heat content of the fat = heat content of water +heat content of the lipids
Let it be Q 
the Q= (mcΔT)_lipids + (mcΔT)_water
total mass of fat  M= 0.63 Kg
Q= heat supplied = 100 W in 5 minutes
ΔT= 20°C
c_lipid= 1700J/(kgoC) 
c_water= 4200J/(kgoC) 
then, 

solving the above equation we get 
m= 0.46 kg 
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg 
 
        
             
        
        
        
To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.
- Like the majority of software engineering initiatives, the ER process begins with gathering user requirements. What information must be retained, what questions must be answered, and what business rules must be implemented (For instance, if the manager column in the DEPARTMENT table is the only column, we have simply committed to having one manager for each department.)
 
- The end result of the E-R modeling procedure is an E-R diagram that can be roughly mechanically transformed into a set of tables. Tables will represent both entities and relationships; entity tables frequently have a single primary key, but the primary key for relationship tables nearly invariably involves numerous characteristics.
 
To know more about  entity AND relationship  visit : brainly.com/question/28232864
#SPJ4
 
        
             
        
        
        
The time of motion of the track star is determined as 0.837 s.
<h3>Time of motion of the track star</h3>
The time of motion of the track star is calculated as follows;
T = (2u sinθ)/g
where;
- T is time of motion
 - g is acceleration due to gravity
 - θ is angle of projection
 
T = (2 x 12 x sin20)/9.8
T = 0.837 s
Learn more about time of motion here: brainly.com/question/2364404
#SPJ1
 
        
             
        
        
        
Answer:
4.25 m/s
Explanation:
Force, F = 22 N 
Time, t = 0.029 s
mass, m = 0.15 kg 
initial velocity of the cue ball, u = 0 
Let v be the final velocity of the cue ball. 
Use newton's second law 
Force = rate of change on momentum 
F = m (v - u) / t 
22 = 0.15 ( v - 0) / 0.029 
v = 4.25 m/s 
Thus, the velocity of cue ball after being struck is 4.25 m/s.