Answer:
P=900KG/M3
U=0.0002 M2/S
RE=PV/U
=900*10/0.0002
=45000000
Explanation:
The Reynold number will be 4.5×10⁷. Reynold's number is found as the ratio of the inertial to the viscous force.
What is density?Density is defined as the mass per unit volume. It is an important parameter in order to understand the fluid and its properties. Its unit is kg/m³.
The mass and density relation is given as
mass = density × volume
The ratio of inertial to viscous force is known as Reynold's number.
[tex]\rm R_E= \frac{\rho u L}{\mu} \\\\ \rm R_E=\frac{900 \times 10}{0.0002} \\\\ R_E=45000000[/tex]
Hence, the Reynold number will be 4.5×10⁷.
To learn more about the density refers to the link;
brainly.com/question/952755
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FOR BRAINLIST HELP PLEASE IS A DCP
A- Causes of the 13t Amendment
B- Reasons for Women's Suffrage
C- Reasons for the Freedmen's Bureau
D- Causes of the Plantation System
Answer:
C
Explanation: Freedmens Bureau provided resources for southerners and newly freed slaves
A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant- 134a at 1.2 MPa and 36°C. Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has
entered the tank and (b) the amount of heat transfer.
Answer:
a) 0.50613
b) 22.639 kJ
Explanation:
From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C
first step : calculate the volume of R-13a ( values gotten from table A-11 )
V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3
next : calculate final specific volume ( v2 )
v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg
a) Calculate the mass of refrigerant that entered the tank
v2 = Vf + x2 * Vfg
v2 = Vf + [ x2 * ( Vg - Vf ) ] ----- ( 1 )
where: Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg ( insert values into equation 1 above )
x2 = ( 0.01652 - 0.0008261 ) / 0.031834
= 0.50613 ( mass of refrigerant that entered tank )
b) Calculate the amount of heat transfer
Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )
uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg
therefore U2 = 164.737 Kj/kg
The mass balance ( me ) = m1 - m2 --- ( 3 )
energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he
therefore Qin = 41.184 - 422.2 + 403.655 = 22.639 kJ
Suppose we are given three boxes, Box A contains 20 light bulbs, of which 10 are defective, Box B contains 15 light bulbs, of which 7 are defective and Box C contains 10 light bulbs, of which 5 are defective. We select a box at random and then draw a light bulb from that box at random. (a) What is the probability that the bulb is defective? (b) What is the probability that the bulb is good?
Answer:
0.49
0.51
Explanation:
Probability that bulb is defective :
Let :
b1 = box 1 ; b2 = box 2 ; b3 = box 3
d = defective
P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))
P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))
P(defective bulb) = 10/60 + 7/45 + 5/30
P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888
= 0.49
P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51