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Pachacha [2.7K]
3 years ago
5

The chart describes four people’s credit histories.

Engineering
2 answers:
icang [17]3 years ago
7 0

Answer:

D). Eesha  pays more than the minimum payment each month.

Explanation:

Eesha would be considered most creditworthy among the given persons as she not only pays on time but also repays more than the minimum amount assigned to pay each month. In order to test the creditworthiness of an individual, his ontime debt paying capability is tested at first followed by the past credit repayment history and the credit score. Except Eesha, all the given candidates have failed to make timely repayment of their debts and hence, they cannot be considered creditworthy.

Alexeev081 [22]3 years ago
5 0

Answer:

D

Explanation:

Hope this helps my dude’s

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Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
3 years ago
Which of the following are examples of engineering controls? Select all that apply.
Neporo4naja [7]

The examples of engineering controls is Biohazard waste containers and Spill clean up kits.

What is engineering controls?

An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.

An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.

Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.

Therefore, the Option C and D is correct.

8 0
2 years ago
What is the best countermeasure against social engineering?
Mkey [24]

Answer:

Hello Monk7294!

Answer:

Employee education

Explanation:

The most important countermeasure for social engineering is employee education. All the employees should be trained to keep confidential data safe. As a part of security education, organizations have to provide timely orientation about their security policy to new employees. The security policy should address the consequences of the breaches.

<em>- I Hope this helps Have an awesome day!</em>

<em>~ Chloe marcus <3</em>

3 0
3 years ago
. In a water cooling tower air enters at a height of 1 m above the ground level with velocity of 20 m/s and leaves the tower at
Zinaida [17]

Answer:

w ( mass flow rate of air ) = 3.16 kg/s

Explanation:

<u>Determine the mass flow rate of air </u>

mass flow rate of water = 1.5 kg/s

Height at which air enters the cooling tower = 1m

velocity of air entering at 1 m = 20 m/s

Height at which air leaves the cooling tower = 7 m

attached below is a detailed solution of the problem

4 0
3 years ago
4.A compression spring has a diameter of 0.67 in and a coil diameter of 0.067 in, with 30 total coils. The spring is to be made
Akimi4 [234]

Answer:

Fatigue factor of safety is 2.0267

Explanation:

Solution is attached below.

4 0
4 years ago
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