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
Answer:
The mass flow rate of cooling water required to cool the refrigerant is
.
Explanation:
A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:



The mass flow rate of the cooling water is now cleared:

Given that
,
,
and
, the mass flow of the cooling water is:


The mass flow rate of cooling water required to cool the refrigerant is
.
Answer:




Explanation:
From the question we are told that:
Zener diode Voltage 
Zener diode Current 
Note

Supply Voltage 
Reduction Percentage 
Generally the equation for Kirchhoff's Voltage Law is mathematically given by



Therefore




Generally the equation for Kirchhoff's Current Law is mathematically given by




Therefore



Answer:
weight is 12.6 N
force is 4.05 N
Explanation:
given data
acceleration = 1.62 m/s²
radius = 1738 km
mass = 10 kg
distance = 1738 km
to find out
weight and force
solution
we apply here weight formula that is
weight = mass × acceleration ...................1
put here value
weight = 10 × 1.26
weight = 12.6 N
and
force = mass × An
force = 10 × 1.62 (1738/ 1738+1738)² = 4.05 N
so force is 4.05 N
Answer:
A. N type impurities
B. P type impurities
Explanation:
A. The impurities contribute free electrons and changing the conducting property of the semi conductor. When a pentavalent impurities in a semi conductor( impurities with five valence electron) , the impurity atom replace some of the semi conductor atoms in the crystal structure where 4 of the valence electron would be involved in bonding of 4 neighbouring semiconductor while leaving the fifth electron to be free(negative charge carrier) which is available for detachment.
B. When a trivalence impurity is added to semiconductor, instead of excess electron, there will be excess hole created by crystals. Reason for this attribute is the trivalence atom will replace some tetra valence semiconductor atom, when three valence electrons of the 3 valence electrons of the trivalent impurity atom make bond with three neighbouring semiconductor which gives rise to lack of electron in the bond of the fourth neighbouring semiconductor which contribute a whole to the crystalline since trivalent impurity contribute excess holes to the crystal of semi conductor, this holes can accept electrons.