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Mumz [18]
3 years ago
6

Technician A says that a tie rod end is a ball and socket joint similar in construction to a ball joint. Technician B says that

tie rod adjustment is use to set
camber. Who is correct?

A. Technician A only

B. Technician B only

C. Technician A and B

D. Neither Technician A and B
Engineering
2 answers:
Stella [2.4K]3 years ago
7 0
It’s A...................
aleksklad [387]3 years ago
5 0

Answer:it is c i think

Explanation:

i have went through this class before

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5. Can you still prepare solutions using available materials at home why​
svetoff [14.1K]
Yes, we can by the proper guidance and proper doing. We can create solution by any materials in the house that you need
7 0
3 years ago
prove that the heat transfer at the constant pressure is given by the enthalpy change during the process​
mihalych1998 [28]

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

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5 0
3 years ago
What are the coventional representative of automation
Oxana [17]

Answer:

ere el merjor 5iyer

Explanation:

yyhh espero ayuder 8 mucho 666

5 0
3 years ago
Explain what a stress-concentration is and how it can be accounted for in the design and analysis of a structural component. Be
Mama L [17]

Answer:

Stress-concentration

        In the machine component the stress become concentrate at the point a particular point and due to this crack formation takes places and after some time some the machine component become fails.High stress concentration means high stress at that point.

It is very important in the design point of view because we have to find out where stress concentration is high and needs to take proper factor of safety to avoid the failure of the machine component.Generally at sharp corner,sudden change and complex shapes stress concentration is high.

3 0
3 years ago
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12
Nina [5.8K]

Answer:

W = - 184.8 kW

Explanation:

Given data:

P_1 = 1.05 bar

T_1 = 300K

\dot V_1 = 84 m^3/min

P_2 = 12 bar

T_2 = 400 K

We know that work is done as

W = - [ Q + \dor m[h_2 - h_1]]

forP_1 = 1.05 bar,  T_1 = 300K

density of air is 1.22 kg/m^3 and h_1 = 300 kJ/kg

for P_2 = 12 bar, T_2 = 400 K

h_2 = 400 kj/kg

\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

8 0
3 years ago
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