Answer:

Explanation:
From the question we are told that:
Velocity of water 
Height=?
Generally, the equation for Water Velocity is mathematically given by

Therefore Height h is given as



The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :


J = 207.6 
So the shear stress at point A is :



Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Answer:
Given that the temperature of the window is below the dew point it will condensate.
Explanation:
A psychrometric chart (like the one attached) will give you the information needed. This chart is for 14.696 psia.
On the bottom horizontal axes you have the dry-bulb temperature, in this case 70°F, going up from this point you can reach the 50% relative humidity curve (red point on chart), going horizontally from this point to the 100% relative humidity you get the dew point temperature (the point at which moisture will condensate) (blue point on chart). In this case the dew point is 50°C. Given that the temperature of the window is below the dew point it will condensate.
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N