The answer is A. Immediately inform her colleague
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
A. Manufacturers rating capacity ↔ 3. Must be marked on all jacks; must not be exceeded
B. Block Used to lift and hold heavy loads, allow them for travel ↔ 1. Place the jack head against this
C. Level surface ↔ 4. Place this under the base of the jack when it's necessary to provide a firm foundation
D. Jack ↔ 2. Used to lift and hold heavy loads, allow them for travel
Explanation:
The manufacturers rating for a jack is labelled on all jacks and should be referenced to compare with the load to be lifted so as to ensure a safe and successful lifting.
In order to lift a load, such as a car, it is required to place the jack on a level surface to provide balance during the lifting task
The head of the jack is placed against the block for lifting heavy objects for proper performance
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432