Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;

The Biot number is given as;

< 0.1, thus apply lumped system approximation to determine the constant time for the process;

The time for the heating process is given as;

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Answer:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
<em>attached below is a detailed solution</em>
Answer:
Output voltage equation is 
Explanation:
Given:
dc gain
dB
Input signal 
Now convert gain,

DC gain at frequency
is given by,



At zero frequency above equation is written as,


Now we write output voltage as input voltage,

Therefore, output voltage equation is 