Answer:
G = $37,805.65
Explanation:
I found this on another site:
475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)
475,000 = 25,000(4.3553) + G(9.6842)
9.6842G = 366,117.50
G = $37,805.65
Answer:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
Answer:
Explanation:
Given data:
initial construction co = 0.286 wt %
concentration at surface position cs = 0 wt %
carbon concentration cx = 0.215 wt%
time = 7 hr

for 0.225% carbon concentration following formula is used

where, erf stand for error function




from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815
from given table



x = 0.002395 mm
Answer: the increase in the external resistor will affect and decrease the current in the circuit.
Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is
E = IR + Ir = I(R + r)........(1)
Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)
I = E/(R + r)
As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.
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