Answer:
t = 6179.1 s = 102.9 min = 1.7 h
Explanation:
The energy provided by the resistance heater must be equal to the energy required to boil the water:
E = ΔQ
ηPt = mH
where.
η = efficiency = 84.5 % = 0.845
P = Power = 2.61 KW = 2610 W
t = time = ?
m = mass of water = 6.03 kg
H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)
<u>t = 6179.1 s = 102.9 min = 1.7 h</u>
When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.
The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.
A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 * 
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.
<h3>What is the mass of casein in wet casein?</h3>
The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg
Mass of water 250 kg
The mass of casein is constant while the moisture content can be changed.
At 12% moisture content;
750 kg = 88%%
100 % = 100 ×750/88 = 852.27 kg
Therefore, the mass of dried casein produced os 852.3 kg.
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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
