Answer:
<em>v</em><em> </em>= T/(2R)
Explanation:
Given
R = radius
T = strength
From Biot - Savart Law
d<em>v</em> = (T/4π)* (d<em>l</em> x <em>r</em>)/r³
Velocity induced at center
<em>v </em>= ∫ (T/4π)* (d<em>l</em> x <em>r</em>)/r³
⇒ <em>v </em>= ∫ (T/4π)* (d<em>l</em> x <em>R</em>)/R³ (<em>k</em>) <em>k</em><em>:</em> unit vector perpendicular to plane of loop
⇒ <em>v </em>= (T/4π)(1/R²) ∫ dl
If l ∈ (0, 2πR)
⇒ <em>v </em>= (T/4π)(1/R²)(2πR) (<em>k</em>) ⇒ <em>v </em>= T/(2R) (<em>k</em>)
Answer:
The idle speed of a running compression should be between 50-75 PSI and that is about half of the static compression.
Explanation:
The Running or Dynamic compression is used to determine how well the cylinder in an engine is absorbing air, reserving it for the proper length of time, and releasing it to the exhaust. The static or cranking compression test is used to check the sealing of the cylinder. Before performing the running compression test, the static compression test is first performed to rule out other issues like bent valves.
The standard value for the static compression is given by;
Compression ratio * 14.7 = Manufacturers Specification
The running compression should always be half of the static compression.
Answer:
Cylindrical
Explanation:
<em>A cylindrical grinder </em><em>is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this.</em><em> Cylindrical grinding</em><em> machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.</em>
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

Replacing the diameter the area results:

Therefore the the stress results:
