Answer:
a)COP=5.01
b)
KW
c)COP=6.01
d)
Explanation:
Given
= -12°C,
=40°C
For refrigeration
We know that Carnot cycle is an ideal cycle that have all reversible process.
So COP of refrigeration is given as follows
,T in Kelvin.

a)COP=5.01
Given that refrigeration effect= 15 KW
We know that 
RE is the refrigeration effect
So
5.01=
b)
KW
For heat pump
So COP of heat pump is given as follows
,T in Kelvin.

c)COP=6.01
In heat pump
Heat rejection at high temperature=heat absorb at low temperature+work in put

Given that
KW
We know that 


d)
Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557
The speed of the car A is 6.05 m/s and the speed of the car B just after the collision 7.65 m/s.
Ma=15 mg , Mb=25mg
Vai=5 m/s vbi=7 m/s
We know coeffecient of restitution
e=|Vaf-Vbf/Vai-Cbi|
0.8=|Vaf-Vbf/5-7|
Vaf-Vbf=1.6
MaVa+mbVb=MaVaf+MbVbf
15*5*25*7=15Vaf+25Vbf
3Vaf+5Vbf=50
sovleving eq 1 and 2
Vbf =6.05 m/s
Vaf=7.65 m/s
The speed of a change in an object's location in any direction. The distance traveled divided by the time required to travel that distance is the definition of speed.Due to its lack of magnitude and merely having a direction, speed is a scalar number. The average speed of an object can be determined if you know the distance traveled and the time it took. Distance times speed is how speed is calculated.
Learn more about speed here:
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