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3 years ago

10

In some companies, workers who increase the quantity or quality of their work receive a. benefits. c. performance bonuses. b. pe

nsion plans. d. orientation. Please select the best answer from the choices provided A B C D

1 answer:

Greeley [361]3 years ago

3 0

Answer:

performance bonuses

Explanation:

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Which of the following activities could be considered unethical?

Zigmanuir [339]

Answer: u slap birds

Explanation: u order cheeseburgers with n cheese

8 0

3 years ago

What is the impedance of a 5μF capacitor at a frequency of 500Hz? What is the impedance of a60mH inductor at this frequency?

choli [55]

Answer:

1) 63.66 ohm

2) 188.49 ohm

Explanation:

Data provided in the question:

Part 1

Capacitance, C = 5μF = 5 × 10⁻⁶ F

Frequency = 500 Hz

Now,

Impedance = $\frac{1}{2\times\pi\times f\times C}$

or

Impedance = $\frac{1}{2\times\pi\times500\times5\times10^{-6}}$

or

Impedance = 63.66 ohm

Part 2

Inductance = 60 mH = 60 × 10⁻³ H

Frequency = 500 Hz

Now,

Impedance for an inductor = 2πfL

thus,

Impedance = 2 × π × 500 × 60 × 10⁻³

= 188.49 ohm

4 0

3 years ago

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0

3 years ago

A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?

Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is $L=100\ cm$

Strain experienced is $\epsilon=82\%=0.82$

Strain is the ratio of the change in length to the original length

$\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm$

Therefore, new length is given by (Considering the load is tensile in nature)

$\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm$

Thus, option (b) is correct.

8 0

3 years ago

A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half it

horsena [70]

Answer:

The resulting pressure is 300 kilopascals.

Explanation:

Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:

$P_{1}\cdot V_{1} = P_{2} \cdot V_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in litres.

If we know that $\frac{V_{1}}{V_{2}} = 2$ and $P_{1} = 150\,kPa$ , then the resulting pressure is:

$P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}$

$P_{2} = 300\,kPa$

The resulting pressure is 300 kilopascals.

6 0

3 years ago