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melamori03 [73]
3 years ago
6

Match the use of the magnetic field to its respective description.​

Engineering
1 answer:
Vesna [10]3 years ago
6 0

oooExplanation:

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Nonshielded cable with a 1.5-inch diameter should have a minimum bending radius of
Flura [38]

Answer:

c is the answer because we have to double the number

6 0
1 year ago
In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subs
NeTakaya

Answer:

For detailed answer of "

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subsea surrounding of S°C. As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of O.S m and wall thickness of 8 mm is used for transporting liquid hydrocarbon at an average temperature of 70°C, and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be 2SO W/m2.K. The subsea surrounding has a temperature of soc and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be ISO W /m2 .K. If the pipeline is made of material with thermal conductivity of 60 W/m.K, by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, (b) determine the inner surface temperature of the pipeline wall, (c) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and (d) determine the heat flux through the outer pipeline surface."

see attachment.

Explanation:

Download pdf
3 0
3 years ago
If the Poisson’s ratio of a 5 mm X 5 mm titanium alloy pin is 0.31 and it is elastically loaded
leonid [27]

The new dimensions of the titanium alloy pin will be that the width is 0.0775 mm and the length is 4.9225m.

<h3>What is Poisson's ratio?</h3>

The Poisson's ratio is the proportion of a material's change in width per unit width to its change in length per unit length due to strain. In order for a stable, isotropic, linear elastic material to have a positive Young's modulus, shear modulus, and bulk modulus, the Poisson's ratio must be between 1.0 and +0.5. Poisson's ratio values for the majority of materials fall between 0.0 and 0.5.

The formula for the longitudinal strain is:

= Change in length / Initial length

Based on the information, the longitudinal strain will be:

= 105 - 100 / 100

= 0.05

Poisson ratio will be illustrated as the change in the width divided by the longitudinal strain. :

0.31 = ∆w/5 / 0.05

∆w = 0.0775 mm

New side length will be the difference in the changes in the dimensions:

= w - ∆w

= 5 - 0.0775

= 4.9225m

Learn more about Poisson on:

brainly.com/question/7879375

#SPJ1

7 0
1 year ago
Two technicians are discussing cylinder- testing. Technician A says that when testing the power level of a specific cylinder usi
konstantin123 [22]

Answer:

The answer is "Both Technician A and Technician B".

Explanation:

The cylinder Testing is intended to assess locomotive inconsistency in CNS rodents, for example, whenever the animal moves within a transparent plastic tube, its preliminary activity is registered as it rises against the stadium wall.

In the given question both technicians are correct because both are reliable ways to check cylinders and the influence of the belief if every pathway has many more advantages than each other.

6 0
3 years ago
Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
Xelga [282]

Answer:

(a) Current density at P is J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

where \textbf{a}_{\rho} and \textbf{a}_{\phi} unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

(b) Total current flowing outward can be calculated by using the relation,

I=\int {\textbf{J} \, \textbf{ds}

where integral is calculated through the circular band given in the question. We can write the integral as below,

I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

4 0
3 years ago
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