Answer:
4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C
Explanation:
we have given 4140 steel contains 0.4% C
we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element
and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel
while in 1045 steel contains 0.45 % c is plain carbon steel
and it do not contain any alloying element
so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C
Sensor/Detectors/Transducers are electrical, opto-electrical, or electronic devices composed of specialty electronics or otherwise sensitive materials, for determining if there is a presence of a particular entity or function. Many vehicles including cars, trains, buses etc. all use sensors to monitor oil temperature and pressure, throttle and steering systems and so many more aspects.
Answer:
a) 70.29 %
b) 37%
Explanation:
percent reduction can be found from:
PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2
= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2
=70.29 %
percent elongation can be found from:
EL =L_f - Lo/Lo*100
= (73.17 -53.3/53.3)*100
= 37%
Answer:
Carpenter's square
Explanation:
The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.
Other hand tools that are used to measure angles are;
- The combination square that allows a user to set both 90° and 45° angles
- A Bevel that allows users to set any angle they like.
- A Protractor that resembles a bevel but its marks are marked in an arc.
- An electromagnetic angle finder which gives a reading according to the measure of the arms adjusted by the user.
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V = 
V = Av = 
V = 
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = 
= 
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = 
v2 = 
v2 = 6.017 m/s
