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Ipatiy [6.2K]
1 year ago
9

joe is an exceptional systems engineer who hopes to work at the techtonic group. he is handicapped and uses a wheelchair. techto

nic does not have wheelchair ramps and some doorways are not wide enough for a wheelchair. the hr department should
Engineering
1 answer:
ASHA 777 [7]1 year ago
5 0

The HR department should install wheelchair ramps and widen the doorways as they don't want to loose an exceptional systems engineer who hopes to work at the techtonic group.

What is a wheelchair ramps?

A wheelchair ramp is an inclined plane that can be installed in addition to or instead of stairs. Ramps make it easier for wheelchair users, as well as people pushing strollers, carts, or other wheeled objects, to enter a building.

The Americans with Disabilities Act requires wheelchair ramps (or other ways for wheelchair users to access a building, such as a wheelchair lift) in new construction for public accommodations in the United States.

A wheelchair ramp can be fixed, semi-fixed, or portable. Permanent ramps are intended to be bolted or otherwise attached to the ground. Semi-permanent ramps are commonly used in the short term and rest on top of the ground or concrete pad.

To know more about wheelchair ramps, visit: brainly.com/question/17395685

#SPJ4

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can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and a
anyanavicka [17]

Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
  • branch 2: i = 21.466∠63.435°; pf = 0.447 leading
  • total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915\times10^{-3}\approx j4.99984\,\Omega

The capacitive reactance is ...

  X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega

<u>Branch 1</u>

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

<u>Branch 2</u>

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

<u>Total current</u>

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

<em>Additional comment</em>

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

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Answer:

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They all represent ultimate improvement.

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If the load parameters are: Vln=600kV, Il=100A (resistive), calculate the source voltage and current when the line is 50Miles (s
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