Answer:
The 5/16 – 24 UNF is stronger because it has more tensile load capacity.
Tensile load capacity for M8 -1.25 = 5670 lb
Tensile load capacity for M8 -1 = 6067 lb
Explanation:
For 5/16 - 18 UNC thread:
D = 0.3125
n = 18
Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18) ^2
= 5243 lb.
Similarly for 5/16 - 24 UNF , only the n value changes to 24
we get the tensile load capacity = 5806.6 lb
Hence the 5/16 – 24 UNF is stronger because it has more tensile load capacity.
For metric Bolts:
We have to consider all values in SI units
Strength = 689 MPa
We get for M8 -1.25:
Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb
For M8 -1:
Tensile load capacity as = 689 X 39.167 = 26986 N = 6067lb
The maximum volume flow rate of water is determined as 0.029 m³/s.
<h3>Power of the pump</h3>
The power of the pump is watt is calculated as follows;
1 hp = 745.69 W
7 hp = ?
= 7 x 745.69 W
= 5,219.83 W
<h3>Mass flow rate of water</h3>
η = mgh/P
mgh = ηP
m = ηP/gh
m = (0.82 x 5,219.83)/(9.8 x 15)
m = 29.12 kg/s
<h3>Maximum volume rate</h3>
V = m/ρ
where;
- ρ is density of water = 1000 kg/m³
V = (29.12)/(1000)
V = 0.029 m³/s
Learn more about volume flow rate here: brainly.com/question/21630019
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Answer:
When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.
Explanation:
When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.
The engine knock problem can be caused due to the following reason
a) When the octane rating of the fuel used is low.
b) The deposition of the carbon around the cylinder walls takes place.
c) The spark plug used in the vehicle is not correct.
Answer:
C. the object is moving forward
Explanation:
A positive slope means position is increasing when time is increasing. Generally, increasing position is "moving forward."
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.
