Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration
State 2 : constant speed
State 3 : deceleration
Total distance : state 1+ state 2+state 3
the average velocity = total distance : total time
Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36
The strain of the bar when subjected to 150 MPa is
Therefore, if the bar elongates by 1.35 mm, then the original length L would be:
or 1.8m