Answer:
Both Technician A and B are correct.
Explanation: Both are correct, but keep note that different color coolants leave different color stains.
Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : 
Standard deviation : 
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula,
, the z-value corresponds to 2.39 will be :-

z-value corresponds to 2.60 will be :-

Using the standard normal table for z, we have
P-value = 

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
Answer:
the pressure gradient in the x direction = -15.48Pa/m
Explanation:
- The concept of partial differentiation was used in the determination of the expression for u and v.
- each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.