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Wittaler [7]
3 years ago
10

A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 83,051 N , and experien

ces an axial elongation of 0.18 mm. Assuming that the deformation is entirely elastic, calculate the elastic modulus of this steel in GPa. Answer Format: X (no decimal places) Unit: GPa

Engineering
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

the answer is attached with required units.

Explanation:

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What is the difference between absolute and gage pressure?
Len [333]

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

P_{ab}=P_{g}+P_{atm}

Here, P_{ab} is absolute pressure, P_{g} is gauge pressure andP_{atm} is atmospheric pressure.

Step2

Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.  

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4 years ago
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3 years ago
Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and req
Genrish500 [490]

Answer:

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Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
3 years ago
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic
DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm

2) The speed of the rotor is the motor speed. The slip is given by:

Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm

3) The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=1*60=60\ Hz

6 0
3 years ago
Hello , how are yall:))))
SVEN [57.7K]

Answer:

eh I'm good hbu?????????

6 0
3 years ago
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