Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
Both of the technicians are correct.
<u>Explanation:</u>
The thrust angle is defined as the angle formed by the imaginary line drawn perpendicular to the rear axis center line. It is used in alignment of the four wheels.
It is also used to determine the direction the front wheels are pointing. While the scrub radius is the intersecting point of the vertical center line of front tires with the imaginary line drawn from the steering knuckles.
Thus both the technicians are saying correct. The thrust angle and scrub radius are used to determine the alignment of wheels, if there is any misalignment in wheels then it needs to make it correct to prevent accidents.
Answer:
Explanation:
Given:
Tooth Number, N = 24
Diametral pitch pd = 12
pitch diameter, d = N/pd = 24/12 = 2in
circular pitch, pc = π/pd = 3.142/12 = 0.2618in
Addendum, a = 1/pd = 1/12 =0.08333in
Dedendum, b = 1.25/pd = 0.10417in
Tooth thickness, t = 0.5pc = 0,5 * 0.2618 = 0.1309in
Clearance, c = 0.25/pd = 0.25/12 = 0.02083in
Answer: The overhead percentage is 7.7%.
Explanation:
We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.
We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.
So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:
OH % = (58 / 758) * 100 = 7.7 %
