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yawa3891 [41]
3 years ago
11

Technician A uses three prong electrical cords when possible.

Engineering
1 answer:
Zolol [24]3 years ago
8 0
A

Correct me if I’m wrong tnx
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Water at 60°F passes through 0.75-in-internal diameter copper tubes at a rate of 1.2 lbm/s. Determine the pumping power per ft
Lelu [443]

Answer:

The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts

Explanation:

See calculation attached.

- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.

- Calculate the cross-sectional flow area.

- Calculate the average velocity of water in the copper tubes.

- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.

8 0
3 years ago
What happens to the electrolyte, during discharging?
lisov135 [29]
During the discharge ions combine with anode to form a compound and release one or more electron.
3 0
3 years ago
A. Briefly define identity theft.
stira [4]

Answer:

Pharming

Phishing

Explanation:

Pharming is used equal like Phishing, because using fake websites, but in this case, is not necessary spam in your email.

Phishing is used like spam, and they pretend to be real companies by email, if you click in the email you will go a supposedly a real website, you put your information like passwords in these websites

You can prevent these techniques using antivirus and you can verify the websites because fake websites don't have the https:// protocol.

A business can prevent this issue using corporatize email and configuring the company network with security protocols and encrypt the information.

3 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
andriy [413]

Answer:

Uper Bound = 175.5 GPa, Lower Bound = 85.26 GPa

Explanation:

Rule of mixtures equations:

For a two-phase composite, modulus of elasticity upper-bound expression is,

E(c)(u) = E(m) x V(m) + E(p) x V(p)

Here, E(m) is the modulus of elasticity of matrix, E(p) is the modulus of elasticity of patriciate phase, E(c) is the modulus of elasticity of composite, V(m) is the volume fraction of matrix and V(p) is the volume fraction of composite.

For a two-phase composite, modulus of elasticity lower-bound expression is,

E(c)(l) = E(m) x E(p)/V(m) x E(p)+V(p) x E(m)

<u>Consider the expression of rule of mixtures for upper-bound and calculate the modulus of elasticity upper-bound.</u>

E(c)(u) = E(m) x V(m) + E(p) x V(p), (1)

<u>Calculate the volume fraction of matrix.</u>

V(m) + V(p) = 1

<u>Substitute 0.35 for V(p).</u>

V(m) + 0.35 = 1

V(m) = 0.65

From equation (1);

<u>Substitute 60 GPa for E(m), 390GPa for E(p), 0.65 for V(m) and 0.35 for V(p).</u>

E(c)(u )= E(m) x V(m) + E(p) x V(p)

E(c)(u) = (60 × 0.65) + (390 × 0.35)

E(c)(u) = 175.5 GPa

The modulus of elasticity upper-bound is 175.5GPa.

The modulus of elasticity of upper-bound can be calculated using the rule of mixtures expression. Since the sum of volume fraction of matrix and volume fraction of composite is equal to one V(m) + V(p) = 1. Substitute the value of volume fraction of matrix as 0.69 and obtain the volume fraction of matrix.

<u>Consider the expression of rule of mixtures for lower-bound and calculate the modulus of elasticity upper-bound.</u>

E(c)(l) = (E(m) x E(p))/ (V(m) x E(p) + V(p) x E(m))

<u>Substitute 60 GPa for E(m), 390GPa for E(p), 0.65 for V(m) and 0.35 for V(p).</u>

E(c)(l) = 60 × 390/(0.65 × 390) +(0.35 × 60)

E(c)(l) = 23400/274.5

E(c)(l) = 85.26 GPa

The modulus of elasticity lower-bound is 85.26 GPa.

8 0
4 years ago
For an applied transformer with a primary winding of 220, a secondary winding of 113 turns, a core of 45 cm2 cross-section and a
Damm [24]

Explanation:

https://www.chegg.com/homework-help/questions-and-answers/iron-core-wish-design-transformer-part-dc-power-supply-moderately-high-current-rating-volt-q53186459?trackid=83dbc34cec2e&strackid=3efc9f324415

8 0
3 years ago
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