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2 years ago

Urgent please help! What are non-ferrous metal and ferrous metal?

1 answer:

8 0

In metallurgy, non-ferrous metals are metals or alloys that do not contain iron in appreciable amounts. Generally more costly than ferrous metals, non-ferrous metals are used because of desirable properties such as low weight, higher conductivity, non-magnetic property or resistance to corrosion

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A pressure gage at the inlet to a gas compressor indicates that the gage pressure is 40.0 kPa. Atmospheric pressure is 1.01 bar.

bonufazy [111]

Answer:

Given

inlet Pga =40kpa = 40000pa

Patm=1.01bar = 1.01 x 100000pa =101000pa

exit Pab= 6.5 (inlet Pab)

But generally, Pab = Patm + Pga

1. the absolute pressure of the gas at the inlet, inlet Pab?

inlet Pab = Patm + inlet Pga

= 101000pa + 40000pa = 141kpa

the absolute pressure of the gas at the inlet, inlet Pab = 141kpa

2. the gage pressure of the gas at the exit? exit Pga?

exit Pab = Patm + exit Pga

exit Pga = exit Pab - Patm

= (6.5 x 141kpa) - 101kpa

= 815.5kpa

the gage pressure of the gas at the exit exit Pga=815.5kpa

5 0

3 years ago

Steam enters an adiabatic turbine at 400◦C, 2 MPa pressure. The turbine has an isentropic efficiency of 0.9. The exit pressure i

pychu [463]

Answer:

Explanation:

Find attached the solution

8 0

3 years ago

Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate

uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae $^{-E/RT}$ ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e $^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}$

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

k = 4.21 * 10⁻³((L/mol)s⁻¹)

k = 4.21 * 10⁻³(L/(mol.s))

3 0

3 years ago

What are the three most common types of relearn procedures?

WITCHER [35]

Answer:

The three types of relearn procedures are auto relearn, stationary and OBD.

Explanation:

In TPMS system, after the direct service like adjustment of air pressure, tire rotation or replacement of sensors etc, is performed then maximum vehicle often needs TPMS system relearn that needs to be performed.

For performing these relearn procedure, there are mainly three types:

auto relearn
stationary relearn
OBD

After applying the relearn process, the TPMS system will again be in proper function.

8 0

3 years ago

4. A banking system provides users with several services:

VLD [36.1K]

A diagram showing a use case diagrams for these requirements is given in the image attached.

<h3>What is system Case diagram?</h3>

A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.

A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.

Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.

Learn more about Case diagram from

brainly.com/question/12975184

#SPJ1

4 0

2 years ago