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Musya8 [376]
3 years ago
15

Which of the following explains why material properties present challenges for engineers?

Engineering
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

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Please help me in this assignment.
xeze [42]
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8 0
3 years ago
Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0
Naily [24]

Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

(a) Find the total current crossing the surface z = 0.1 m in the az direction.

(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.

(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

Answer:

a. -39.8μA

b. -15.81mC/m³

c. 29.05m/s

Explanation:

Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

J = 0.

a. Total current is calculated by.

J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

π = 22/7

φdza = 10^-6

z = 0.1

Total current

= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= −39.7543477278310

= -39.8μA

b. Calculating velocity charge density at (ρv)

Density (J) = ρv * V

Where J = Density = -10^6 * z^1.5

V = 2 * 10^6

z = 0.1

Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)

ρv = -0.1^1.5/(2)

ρv = -0.015811388300841

ρv = -0.01581 --------- Approximated

ρv = -15.81mC/m³

c. Calculating Velocity

Velocity = J/V

Where Velocity Charge Density = -2000 C/m3

Where J = -10^6 * z^1.5

z = 0.15

J = -10^6 * 0.15^1.5

J = -58094.75019311125

Velocity = -58094.75019311125/-2000

Velocity = 29.047375096555625m/s

Velocity = 29.05m/s

8 0
3 years ago
Steam enters an adiabatic turbine at 8 MPa and 500C with a mass flow rate of 3
Vinil7 [7]

Answer:

a)temperature=69.1C

b)3054Kw

Explanation:

Hello!

To solve this problem follow the steps below, the complete procedure is in the attached image

1. draw a complete outline of the problem

2. to find the temperature at the turbine exit  use termodinamic tables to find the saturation temperature at 30kPa

note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa

4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.

Note: Remember that for a turbine with a single input and output, the power is calculated as the product of the mass flow and the difference in enthalpies.

5. Find the real power of the turbine

3 0
3 years ago
Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?
lianna [129]

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents  a crystal system  which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have  three axes which have only two of its axes equal in length.

5 0
3 years ago
Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
vodomira [7]

Answer:

I = 0.625 A

Explanation:

Given that,

Power of the light bulb, P = 75 W

Voltage of the circuit, V = 120 V

We need to find the current flowing through it. We know that, Power is given by :

P=V\times I

I is the electric current

I=\dfrac{P}{V}\\\\I=\dfrac{75\ W}{120\ V}\\\\I=0.625\ A

So, the current is 0.625 A.

5 0
2 years ago
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