Fact
Explanation:
<em>If </em><em>wrong </em><em>Im </em><em>sorry </em><em>Correct </em><em>me </em><em>In </em><em>The</em><em> </em><em>c</em><em>om</em><em>m</em><em>ent </em><em>please </em>
Gas turbines extract the energy from combustion gas
and heat engines convert thermal and chemical energy to mechanical energy
gas is considered a chemical so i'm pretty sure it is considered one
Answer:
(a) ΔU = 125 kJ
(b) ΔU = -110 kJ
Explanation:
<em>(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?</em>
<em />
The work is done to the system so w = 150 kJ.
The heat is released by the system so q = -25 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -25 kJ + 150 kJ = 125 kJ
<em>(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?</em>
<em />
The work is done by the system so w = -100 kJ.
The heat is released by the system so q = -10 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -10 kJ - 100 kJ = -110 kJ
Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
Answer:
T = 
Explanation:
Given,
Initial Velocity, u=0
As platoon is moving with constant velocity v,
Final Velocity, v=v
The vehicle starts from 0 to v at constant acceleration of a,
Relevent expressions:
v=u+at...........................(1)
v2=u2+2as.....................(2)
= 2aS, as S = 
,
= 2a,
From(1)
v=at
Hence
(at)^2=2a
= 
T =
This is the final expression for time.
