Answer:
increases by a factor of 6.
Explanation:
Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:
Initial flow rate = area * velocity = A * V = AV m³/s
The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:
Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate
Hence, the volume flow rate of the water passing through it increases by a factor of 6.
Answer:
Check the explanation
Explanation:
Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

1x0.723x
=3x0.780x
⇒
= 426.4 °K
The initail volumes of the gases can be determined by the ideal gas equation of state,
=
= 0.201
The equilibrium pressure of the gases can also be obtained by the ideal gas equation

= 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4
(0.201+1.275)
= 246.67 KPa = 2.47 bar
The pressure difference across the sensor housing will be "95 kPa".
According to the question, the values are:
Altitude,
Speed,
Pressure,
The temperature will be:
→ ![T = 15.04-[0.00649(9874)]](https://tex.z-dn.net/?f=T%20%3D%2015.04-%5B0.00649%289874%29%5D)
→ 
→ 
now,
→ ![P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}](https://tex.z-dn.net/?f=P_o%20%3D%20101.29%5B%5Cfrac%7B%28-49.042%2B273.1%29%7D%7B288.08%7D%20%5D%5E%7B%285.256%29%7D)
→
hence,
→ The pressure differential will be:
= 
= 
Thus the above solution is correct.
Learn more about pressure difference here:
brainly.com/question/15732832
Answer:
Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.
Hope it help you friend
Explanation:
Sum of forces in the x direction:
∑Fx = ma
Rx − 250 N = 0
Rx = 250 N
Sum of forces in the y direction:
∑Fy = ma
Ry − 120 N − 300 N = 0
Ry = 420 N
Sum of forces in the z direction:
∑Fz = ma
Rz − 50 N = 0
Rz = 50 N
Sum of moments about the x axis:
∑τx = Iα
Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0
Mx = 22 Nm
Sum of moments about the y axis:
∑τy = Iα
My = 0 Nm
Sum of moments about the z axis:
∑τz = Iα
Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0
Mz = -30.8 Nm